MAT4 : Introduction to Applied Linear Algebra Mike Newman fall 7 9. Projections introduction One reason to consider projections is to understand approximate solutions to linear systems. A common example of this is to find the best-fit line to a given set of points: this is the regression line. Problem 9.. Find the best line that approximates the three points ( ) ( ) ( ). Situations like this arise in any context where it is felt that in some sense the value of some parameter (x) should determine the value of some outcome (y) but where there are other sources of variability in the data. So an experimenter measures outcomes (y) for different values of some parameter (x) with the goal of determining the relationship between them. Some examples: one might wish to determined the gravitational constant g at a particular location and use the time taken for an object to fall m to determine g. The actual times will vary due to air turbulence measurement error etc but on average should correlate with the true values of g. Problem 9. is of course somewhat oversimplified: three data points don t really indicate much. But it is small enough that one could plot the three points and try and fit a line as close as possible. This can be seen as an application of projection in a vector space; in fact a projection in a highdimensional vector space. Projections have a number of other uses so we start by establishing a bit of theory before returning to our application above. scalar product The (standard) scalar product of two vectors in R n is defined as follows. x y = x t y = x y + x y + + x n y n Some useful geometrical facts about the scalar product. Proposition 9.. For x y = x t y we have the following. The Euclidean length of a vector x is x = x x. We sometimes call this the norm of a vector. In two dimensions this is the Pythagoras theorem. Furthermore x = if and only if x x = if and only if z =. Two vectors x and y are orthogonal if the angle between x and y is 9 ; we write x y. We have x y if and only if x y =. More generally if we let θ be the angle between x and y then we have x y = x y cos θ. Problem 9.. If we draw x and y there are two angles that could be considered the angle between them. Furthermore we could measure this angle in two different directions so it is unclear if this angle is positive or negative. Explain how these ambiguities do not affect the expression x y cos θ These notes are intended for students in mike s MAT4. For other uses please say hi to mnewman@uottawa.ca. 86
There are also useful algebraic properties of the scalar product. Proposition 9.4. For x y R n and a b R we have the following. x y = y x x (au + bv) = ax u + bx v Proof. We leave this as an exercise using known properties of matrix multiplication. The first property of Proposition 9.4 is called commutativity of the scalar product; the second is called linearity of the scalar product. orthogonal projection The (orthogonal) projection of a vector x on to a vector y is the part of x that is in the direction of y. In order to make this precise we decompose x into two parts: one in the direction of y and the other perpendicular to y. x v y u In this picture the vector u represents the projection of x on to y. We write u = proj y (x). The vector v represents the part of x that is orthogonal to y. We have x = u+v. Knowing the projection u we could find the orthogonal part as v = x u. Problem 9.5. Is proj y (x) = proj x (y)? We can calculate the projection using the following formula. Proposition 9.6. The projection of x on to y is proj y (x) = y x y yy. In particular if we let u = proj y (x) = y x y yy and v = x u then v is parallel to y x is orthogonal to y and x = u + v. Proof. We see that u is parallel to vy because it is a scalar multiple of y. By definition we have x = u + v. So we prove that v is orthogonal to y. ( y v = y t x y x ) y y y = y x y x y y y y = y x y x = Note that we used Proposition 9.4 in simplifying the expression. Example 9.7. Let x = and y = 6. Calculate the projection of x on to y. 87
Solution. We calculate the projection as proj y (x) = 6 = 4 8 We can then calculate the rest of x that is to say the part orthogonal to y as follows. x proj y (x) = 6 8 8 We obtain a decomposition of x in two parts: the first in the direction of y (this is proj y (x)) and the second is orthogonal to y. 6/ 4/ 4 8 8 = + + 6 4/ 6/ 8 Problem 9.8. Check that the decomposition in the previous example has the right directions: the first vector should be a multiple of y and the second should be perpendicular to y. Problem 9.9. Let x = and y =. Calculate proj y (x) and give a decomposition of x into two vectors: one parallel to y and the other orthogonal to y. Also calculate proj x (y) and give a decomposition of x into two vectors: one parallel to x and the other orthogonal to x. projection onto subspaces We can think of Proposition 9.6 in a slightly more general way. We are given a vector x and a subspace U and we want to express x = u + vv where u U and v is perpendicular to U. Putting aside the fact that we don t (yet) have an idea of what it might mean for a vector to be perpendicular to a subspace we can think of Proposition 9.6 as corresponding to the case where U is the span of a single vector y. In this case at least it seems reasonable to say that v is perpendicular to U exactly when it is perpendicular to y. This is in fact the real notion of projection that we want; we will see that Proposition 9.6 essentially extends to projection onto subspaces but we need a bit of work first. In the meantime and inspired by the thought that projection onto y might be the same as projection onto the space spanned by y we notice the following. Proposition 9.. Let z = γy where neither y not z are the zero vector. Then proj y (x) = proj z (x). bases (review) Let B = {v v v k } be a set of vectors in a vector space V. We say that B spans V (or B is a spanning set for V ) if every vector in V can be written as a linear combination of the elements of B. We say that B is linearly independent if the only linear combination of the elements of B that give is the trivial one. In other words B is linearly independent if α v + α v + + α k v k = = α = α = = α k = 88
A basis of a vector space V is a set of vectors in V that is linearly independent and spanning. We have already seen examples in this course. We found bases for eigenspaces of a matrix. For a diagonalizable matrix we saw that taking the union of the bases for each eigenspace we get a basis for R n. We called this an eigenbasis of R n. This was useful in studying dynamical systems. A basis is useful because it gives a unique expression for every vector. Theorem 9.. Let B = {v v v k } be a set of vectors in some vector space V. Then B is a basis if and only if for every vector x in V there exists unique values α α α k that give x = α v + α v + + α k v k. These values α i in Theorem 9. are the coordinates of x with respect to the basis B. Proof. Assume B is a basis. We must show that any x can be uniquely expressed in terms of B. First of all since B spans V there certainly are some numbers α α k such that x = α v + α v + + α k v k. Now assume this expression is not unique that is Then x = α v + α v + + α k v k = β v + β v + + β k v k = x x = (α v + α v + + α k v k ) (β v + β v + + β k v k ) = (α β ) v + (α β ) v + + (α k But B is linearly independent so we must have α i β i = for each i which means that α i = β i and there is exactly one way to express x as a linear combination of the elements of B. Now assume that for every vector x V there is exactly one way to express x as a linear combination of the elements of B. In particular there is some way to express x so B spans V. So for x = there are constants such that = γ v + γ v + + γ k v k. But we know that = v + v + + v k so we must have γ i = for each i which shows that B is linearly independent. { { Example 9.. Check that B = and B } = are both bases. Then find the } coordinates of with respect to both of these. 4 Solution. In order to check that these are both bases we need to check that they are linearly independent and that they span R. Alternatively we need to check that every vector x can be uniquely expressed in terms of the elements of the alleged basis. For B we need to show that for every vector x R there is a unique solution to x α + α = x But this is true if and only if α = x and α = x. So there is a unique solution for every x. For B we want a unique solution to α + α = x x Right about now we should admit to an abuse of notation. When we are thinking of the coordinates of a vector with respect to a basis we are thinking of the basis as an ordered set (sometimes called a tuple). In other words we need to have a way to tell which vector is first so we know which elements α corresponds to and so on for the second third etc. This doesn t usually cause any problem but note that a when we say that a basis is a set of vectors we really mean a set with an ordering. 89
We would consider the following augmented matrix. x x This will have a solution for every x if it has a pivot in every row; it will have a unique solution if it has a pivot in every column (no free variables). So we conclude that a set of vectors is a basis for R n if and only if when we put these vectors as columns in a matrix we have a pivot in every row and column. We leave it as an exercise to apply this to the example above. Now to find coordinates. With respect to B we are looking for numbers α α such that 4 = α + α Here we can see directly that α = and α = 4. With respect to B we are looking for numbers α α such that 4 = α + α As before we might switch to an augmented matrix perspective. 4 4 The coordinates are then α = and α = 4. We notice that a vector can have different coordinates with respect to different bases. In fact we might notice that B and B have exactly one element in common but it is the coordinate of that element that changed. So when we speak of the coordinates of a vector with respect to a basis it is with respect to the whole basis. There are some other results about bases we will find useful. Theorem 9.. Let B and B be bases for a vector space. Then B = B. We say that the dimension of a vector space is the size of a basis. By the previous theorem this is well-defined. In fact we can say a little more than Theorem 9.. Theorem 9.4. Let V be a vector space of dimension n and B a set of vectors in V. If any two of the following is true then so is the third. B is linearly independent B spans V B = n In particular B is a basis of V. orthogonal bases Orthogonality is not just a geometric property: it is an algebraic one also. A set B is an orthogonal set if every pair of vectors in B are orthogonal and it does not contain the zero vector. 9
Theorem 9.5. Let {u u u k } be an orthogonal set. Then {u u u k } is linearly independent. The converse is not true: a linearly independent set is not necessarily orthogonal. Proof. Assume {u u u k } is an orthogonal set and that = α u + α u + + α k u k Now we take the scalar product of each side with u i for some i k. u i = u i (α u + α u + + α k u k ) = α u i u + α u i u + + α k u i u k = + + + α i u i u i + + + = α i We used Proposition 9.4 to distribute the product; we used the fact that u i and Proposition 9. to divide by u i u i at the end. An important consequence of this is that if we have n (nonzero) orthogonal vectors in R n then using Theorem 9.5 and Theorem 9.4 we have a basis for R n. We say that a basis whose elements are pairwise orthogonal is an orthogonal basis. Similarly if we have k orthogonal vectors in a subspace U of dimension k we would have an orthogonal basis for U. Problem 9.6. Let B be an orthogonal set pf vectors from R n. Show that it is an orthogonal basis for the subspace spanned by B. { { Problem 9.7. Verify that is an orthogonal basis for R }. Verify that is } not an orthogonal basis for R. An orthonormal basis is an orthogonal basis where each vector is of length (norm) equal to. We can get an orthonormal basis from an orthogonal one by dividing each by its norm. Example 9.8. Verify that {u u } = is an orthogonal basis for the subspace spanned by these two vectors. Then transform it into an orthonormal basis. Solution. We check that u u = ()( ) + ()() + ( )() = which means that they are orthogonal and hence (Theorem 9.5) linearly independent. They certainly span the space that they span so they are a basis for that subspace. We calculate their norms. u = () + () + ( ) = u = ( ) + () + () = So we have an orthonormal basis (for the space spanned by the two vectors). { } u u = / / / / / Problem 9.9. In an orthonormal basis B the scalar product of two different vectors of B is and the scalar product of a vector of B with itself is. Explain why this is true. This can be understood as follows. 9
Proposition 9.. Let B = {u u k } be an orthonormal basis for some subspace of R n (which might be R n ). If we define the matrix B by B ij = u i u j then B = I. Alternatively if we let Q be the matrix whose columns are the vectors of B then Q t Q = I. A square matrix Q with the property that Q t Q is said to be an orthogonal matrix. Even if Q is not square it is still a nice matrix: see Exercise 9.8. orthogonal bases and projections We would like to know how to compute projections in an efficient manner. Orthogonal bases accomplish this perfectly. Our immediate goal is to obtain a formula for (orthogonal) projection onto a subspace. We will see that this is essentially the same thing as the projection of a vector onto a vector. Theorem 9.. Let U be a subspace of R n (which might be all of R n ) and let {u u u k } be an orthogonal basis for U. For any vector x U we have the following. x = u x u + u x u + + u k x u k u u u u u k u k In other words x is equal to the sum of the projections of x onto each vector in the orthonormal basis. Proof. We know from Theorem 9. that there are unique numbers α i such that x = α u + α u + + α k u k Now we take the scalar product of each side with u i for some i k. u i x = u i (α u + α u + + α k u k ) u i x = α u i u + α u i u + + α k u i u k u i x = + + + α i u i u i + + + u i x u i u i = α i Which means the α i are exactly what they were claimed to be. It is worth comparing this to the proof of Theorem 9.5. We restate Theorem 9. in terms of the coordinates although this is little more than a repetition. Corollary 9.. Let U be a subspace of R n and let {u u k } be an orthogonal basis for U. Then the coordinates of x with respect to this basis are α i = u i x u i u i. Recall our previous goal of wanting to understand how to project onto a subspace. So give a vector x R n and a subspace U of R n we want to find a vector u in U and a vector v orthogonal to U such that x = u + v. 9
For v to be orthogonal to a subspace means that v is orthogonal to each vector in U. We define the orthogonal complement of U written U as the set of all vectors of R n that are orthogonal to every vector of U. Specifically U = {y y x x U} Example 9.. In fact you might have seen an example of this in mat4. Let A be some m n matrix of rank k and denote by U its rowspace. So U is the set of all vectors that are linear combinations of the rows of A. Then U is exactly the kernel (null space) of A. That is U is the set of all vectors x where Ax =. You might recall that the dimension of the rowspace is the rank of A and the dimension of the kernel is the number of free variables of A meaning that dim(u) = k and dim(u ) = n k. In fact this is true for orthogonal complements in general. The theory of orthogonal subspaces gives the following. Theorem 9.4. Let U be a subspace of R n and let U be its orthogonal complement. Then U U = {}. dim(u) + dim(u ) = n. Every vector x can be written uniquely as x = u + v where x U and v U. This is exactly the idea of orthogonal projection on to a subspace. This might make more sense in terms of orthogonal bases. Proposition 9.5. Let U be a subspace of R n and U its orthogonal complement. If B is any basis of U and B is any basis of U then B B is a basis of R n. Furthermore if B and B are both orthogonal bases then so is B B. Proof. Let B = {u u k } and B = {u u n k }. This requires that B + B = n but we know that is true from Theorem 9.4. We show that B B is linearly independent. Assume that Then we have = α u + + α k u k + β u + + β n k u n k α u + + α k u k = (β u + + β n k u n k ) This means that α u + + α k u k is in U and in U and so α u + + α k u k = (see Exercise 9.). But since B is linearly independent the only way this linear combination can be zero is if each of the α i =. Also β u + + β n k u n k is in U and U and by an analogous argument each of the β i must also be zero. If B and B are both orthogonal bases then u i u j whenever i j k or whenever k + i j n. If i k and k + j n then u i U and u j U and so u i u j ; similarly if j k and k + i n. We will see that we can find orthogonal bases for any subspace; and in fact we will see a practical method for extending a basis B of U into a basis B B of R n where B is a basis for U. Putting aside for the moment how exactly we will do this we see what it is good for. 9
Theorem 9.6. Let {u u u k } be an orthogonal basis for U and {v v n k } an orthogonal basis for U. Then we have u U and v U and x = u + v where u = x u u u u + x u u u u + + x u k u k u k u k v = x v v v v + x v v v v + + x v n k v n k v n k v n k Furthermore u and v are the only vectors such that u U v U and x = u + v. That is they do not depend on the particular choice of orthogonal bases. Proof. Given the two orthogonal bases the vectors u and v are completely determined and using Theorem 9. applied to R n we see that u + v = x. On the other hand assume that u U and v U with u + v = x. Then u + x = x = u + v. Subtracting we see that u u = v v. But then u u is in both U and U and so by Exercise 9. we have u u = or u = u. Similarly v = v. We define the projection of x onto a subspace U of R n as proj U (x) = x u u u u + x u u u u + + x u k u k u k u k where {u u k } is an orthogonal basis for U. We repeat that even though this definition is given in terms of a particular basis for U the vector proj U (x) does not depend on the choice of basis. Example 9.7. Let B = and B = 4. Show that these are both orthogonal sets and hence orthogonal bases for the subspaces they span. Show that these two spaces are orthogonal. Considering (somewhat arbitrarily) the first as U and the second as U calculate proj U (x) and proj U (x) for x = t. Solution. We first check that B and B are indeed orthogonal (exercise). We also check that each of the vectors of B is orthogonal to each of the vectors of B ; combining this with Exercise 9.7 we see that the two subspaces are indeed orthogonal as claimed. Note that Proposition 9.5 tells us that the union of these two must be an orthogonal basis for R 4. Now for the projections. / proj U (x) = 9 + = / / / proj U (x) = 6 8 4 + = / 4/ The sum of these two projections is indeed x. So in fact we could have calculated one of them and obtained the other by subtraction. Problem 9.8. For the bases of Example 9.7 find proj U (y) and proj U (y) for y = t. Then calculate y proj U (y) and compare. 94
There are two technical questions. Given a subspace how to find an orthogonal basis? Given a subspace how to find a basis for the orthogonal complement? Gram-Schmidt We are inspired by the first picture of the projection of x on to y. The vectors x and y form a basis for R but not an orthogonal one. But if we consider v = x proj y (x) we see that y and v form an orthogonal basis for R. The idea is to construct the orthogonal basis one vector at a time at each step subtracting from the next vector the projections of it onto the vectors we already have. Algorithm 9.9. Let S = {v v v k } be a set of vectors that span a subspace U of R n. We will build orthogonal basis as follows. We start with B = {} and we repeat the following steps until all vectors in S have been considered.. Choose a vector x from S.. Subtract from x its projection onto each element of B.. If the result is nonzero put it into the set B. At the end the set B will be an orthogonal basis for the space spanned by S. Example 9.. The vectors this subspace. 5 span some subspace. Find an orthogonal basis for Solution. We start with an empty orthogonal set B = {}. We choose some vector of S say We subtract its projections onto all the vectors currently in B. There { is nothing in B (yet!) so } nothing to subtract. We put the result into B. Now we have B =. We repeat! Now we choose some other vector from S say currently in B. We put this result into B. Now we have B = {. We subtract its projections onto the vectors = 6 6 = }.. 95
Now we choose the only remaining vector in S namely vectors currently in B. 5 5 { Now we have B = 5 final B is an orthogonal basis for the space spanned by S. 5. We subtract its projections onto the 5 = 6 6 4 = }. There are no more vectors in S so we are done. The It is important to note that when we are subtracting the projections we are considering projections on to the new vectors up to now not the originals. In other words we project on to the vectors we put in B not the vectors we took out of S. Another observation that is perhaps useful is that the final vectors are orthogonal hence linearly independent. So in Example 9. we found a linearly independent set that spanned the span of S. This means that the span of S is of dimension and so (using Theorem 9.4) we have proved that S is actually a basis. Were it not to have been a basis we would have gotten the vector zero at some step (After subtracting the projections) which we would not have put into B. Problem 9.. In Example 9. we chose the vectors in a particular order. Repeat this example but with a different order. Do we still get an orthogonal basis? Is it the same basis? Problem 9.. In Example 9. we found a basis of three orthogonal vectors. Verify directly that is also an orthogonal set. This little trick is sometimes useful. It says that we can multiply a vector in an orthogonal set by a nonzero constant without destroying the orthogonality. It s not the same basis but it s still an orthogonal basis. This can be especially handy if part way through we start getting ugly fractions: we can rescale the vector before adding it into B. finding the orthogonal complement In some sense this section is optional. Give a subspace U and a vector x we have proj U (x) = x proj U (x). So we don t really need to find a basis for U in order to find proj U (x). But we include the following anyway for completeness and to show that there are other uses for row-reduction. In fact we ll give two methods. Algorithm 9.. Consider a set {v v v k } that spans some subspace U of R n. We put these vectors as columns of a matrix A and we apply Gauss-Jordan elimination to the augmented matrix A I. The result is R B. The columns of R with pivots indicate which columns of the original form a basis for U. The rows of B corresponding to zero rows of R give a basis for U. If we combine the basis of U with the basis of U we obtain a basis of R n (Proposition 9.5). We have completed the basis of U to a basis of R n. 96
Proof Sketch. The method of Algorithm 9. is based on elementary matrices. Every row operation on a matrix is equivalent to multiplying that matrix by an invertible matrix on the left. So we can think of the entire row reduction as multiplying on the left by a single invertible matrix (which is a product of many elementary matrices). Call this matrix E. Then the final augmented matrix after the row-reduction is EA I = EA E. A zero row to the left of the vertical bar is a zero row in EA. But a zero row in EA means that the corresponding row of E is orthogonal to every column of A. Those are the rows that span U. Since the matrix E is to be found to the right or the vertical bar then the corresponding rows form the basis for U. The use of elementary matrices in this proof is closely related to what is happening in Exercise 8.. In that question you were being asked to find the first row of the matrix E; where E corresponds to the set of row operations that transformed one tableau into the other. In Algorithm 9. we are considering this as an augmented matrix so we only care about pivots to the left of the vertical bar. In other words we are really row-reducing A but the I is coming along for the ride. Furthermore we only need to know where the pivots to the left of the vertical bar are we do not need to go any further than that. Example 9.4. A subspace U is spanned by the vectors. Find a basis for U. 4 Solution. We row reduce the following. 4 To the left of the vertical bar we see pivots in the first and second columns. So we take the first and second columns of A. To the left of the vertical bar we see that the third and fourth rows are zero. So we take the third and fourth rows of B as a basis for U. Combining the two gives a basis for R 4. basis of U: basis of U : basis of R 4 : Problem 9.5. Check in Example 9.4 that each vector in the basis for U is orthogonal to each vector in the basis for U. Problem 9.6. In Example 9.4 verify that the basis for U is not orthogonal. Transform is into an orthogonal basis using Algorithm 9.9. Do the same for the basis for U. Verify that these two are indeed orthogonal bases for U and U and that their union is an orthogonal basis for R 4. Why was it not necessary to apply Algorithm 9.9 with four vectors in order to get an orthogonal basis for R 4? What happens if you take the basis for R 4 from Example 9.4 and apply Algorithm 9.9 to this set? 97
We have another method based on the fact that if U is the row space of a matrix then U is the kernel of that matrix. Algorithm 9.7. Consider a set {v v v k } that spans some subspace U of R n. We put these vectors as rows of a matrix Aand row-reduce it. The rows with pivots indicate the rows of the original matrix A that form a basis of U. Since the orthogonal complement of a row space is a kernel then a basis for ker(a) is a basis for U. In other words a basis for U is exactly a basis for the solutions to Ax =. Proof Sketch. A vector x ker(a) is such that Ax = which means that x is orthogonal to every row of A. So ker(a) U. Now we know that the dimension of ker(a) plus the dimension of the rowspace is n the number of columns. But since U is a subspace of R n then Theorem 9.4 tells us that dim(u ) = dim(ker(a)). If ker(a) U then there is a vector in U that is not in ker(a). But then this vector is not in the span of ker(a) so dim(u ) > dim(ker(a)) which is a contradiction. Example 9.8. A subspace U is spanned by the vectors. Find a basis for U. 4 Solution. We row reduce the following. 4 We have two free variables so we will have two vectors in the basis for the kernel. For each free variable we set it to and all other free variables to. x = x 4 = x = x 4 = basis of U: basis of U : basis of R 4 : exercises. Prove Proposition 9.4 using things you know about matrix multiplication.. Prove Proposition 9. using the formula for projection from Proposition 9.6.. Find two different bases B and B for a vector space V and a vector x V such that the coordinates of x with respect to B and B are the same. 98
4. Let B and B be bases for some vector space V. Assume that for every x V the coordinates of x with respect to B and B are the same. Prove that B = B. (hint: what if x B?) { a 5. a) Find all values a and b such that is an orthogonal set. b} b) Find all values a b c d such that a b c d is an orthogonal set. 6. Consider the space U spanned by B = and the vector x = a) Verify that B is an orthogonal set. b) Determine proj u (x) for each vector u B. c) Considering the sum of the three projections is x U? (hint: Theorem 9.) 7. Let B and B be two orthogonal sets. Show that if u is in the span of B and v is in the span of B then u v. 8. Let Q be some matrix (not necessarily square). Show that Q t Q = I if and only if the columns of Q form an orthonormal basis for the subspace that they span. 9. Let U be a subspace of R n. Show that U (U ). Using Theorem 9.4 show that U = (U ). (The former is true in every vector space; the latter can be false in infinitedimensional vector spaces.). For U a subspace of R n show that U U = {}. (hint: let x U U and notice that x x = ). The set is a basis for a subspace U. a) Using Gram-Schmidt (Algorithm 9.9) give an orthogonal basis for U. b) Adjust this to an orthonormal basis.. (not really but we haven t explicitly talked about the subspace test in this course although you would have seen it in mat4) Let U be a subspace of R n. Using the subspace test show that U is a subspace of R n. (This wasn t mentioned explicitly in the text above but is necessary to know to be sure that things like a basis for U make sense.). Let U be the subspace of R spanned by and the vector x = a) Apply the Gram-Schmidt algorithm to find an orthogonal basis for U. b) Redo Gram-Schmidt with the vectors chosen in the other order to find a seceond orthogonal basis for U. c) Find proj U (x). Do this twice using each of your orthogonal bases. Verify that you get the same result. 4. Consider the set S = 4. a) Show that S is an orthogonal set. b) Apply the Gram-Schmidt method (Algorithm 9.9) to S. Explain what happens. Does this seem reasonable? 99
5. Let U be the space spanned by. a) Find a (nonzero) vector orthogonal to U (you can use Algorithm 9. or Algorithm 9.7 but you shouldn t really need either). Using Theorem 9.4 explain why your vector is a basis for U. b) Given x = a b c d t find proju (x) and proj U (x). What happens? Does this seem reasonable? 4 6. Let U be the space spanned by 6. a) Find an orthogonal basis for U. b) For each vector in the standard basis for R 4 find its projection on to U. c) For each vector in the standard basis for R 4 find its projection on to U. 7. Let U be the set of points in R that satisfy ax + bx + cx =. For convenience assume that a. a) Since U = ker( a b c ) we see that U is a subspace of R. Show that a basis for U is c b a. a b) Using Algorithm 9. give a basis for U.