Determinant and Trace

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Bertha McCoy
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1 Determinant an Trace Area an mappings from the plane to itself: Recall that in the last set of notes we foun a linear mapping to take the unit square S = {, y } to any parallelogram P with one corner at the origin. We can write the parallelogram P as P = {v + yw :, y }, where v an w are the two vectors which form the eges of P starting at the origin (, ). Then we can write the linear transformation as ([ ) [ v w T = v + yw, [T =, y v 2 w 2 where v = (v, v 2 ) an w = (w, w 2 ) in components. Notice that the mapping T is invertile precisely when it oes not collapse S own to a line segment (or a point), which happens precisely when the area of the parallelogram P is non-zero. Also, recall that we sai T is invertile precisely when its eterminant et([t ) = v w 2 v 2 w 2 is nonzero. We have et([t ) T invertile Area(P ). We ll see net that et([t ) is the area of P, up to a sign. This is easiest to see with the shear map we eamine in the last set of notes. Start with the sheer map T whose matri representation is [ [T =. (In the earlier set of notes we wrote the entry in the upper right corner of [T as a, ut it will turn out to e convenient to call it for our later iscussion.) In this case, T maps the unit [ square S = {, y } to the parallelogram P spanne y the two vectors v = an [ w = ; in other wors, T (S) = P = {v + yw :, y } = We reprouce a picture here: { [ [ + y } :, y. Area = (, ) Area = (, ) We alreay know that the unit square S has area, ut let s see that P also has area. The area of a parallelogram is equal to its ase times its height, an the height an ase of P are oth, so the area of P is =. On the other han, ([ ) et([t ) = et = = = Area(P ).

2 Now we can rescale the sheer T y a in the horizontal irection an y an vertical irection, to have something more general. This time we have [ a [T =, an T (S) = P = {v + yw :, y } = an the picture looks like { [ a [ + y } :, y, (, ) Area = (a, ) Area = a (In this particular picture a = /2 an = 2, ut this choice of scaling factors is not important.) This time the height of the parallelogram P is while its ase is a, so Area(P ) = ase height = a. Again, we have ([ ) et([t ) = a et = a = a = Area(P ). Notice that the asolute value here is necessary, ecause a an coul have opposite signs. [ a Now that we know et([t ) gives the area of the image of the unit square if [T =, it s not too har to see this is true for any linear map. We ll first nee a technical fact. Eercise: Prove et(ab) = et(a) et(b) for 2 2 matrices A an B. (Really, just multiply it out.) Notice that this means et(ab) = et(ba) for any pair of 2 2 matrices. Eercise: Show that for any angle θ we have ([ ) cos θ sin θ et([r θ ) = et =. sin θ cos θ Now let T : R 2 R 2 e a linear mapping of the plane to itself, an suppose [ a [T =. c [ [ [ [ a This means T (e ) = an T (e c 2 ) =, where e = an e 2 = as efore. [ a Now, the vector T (e ) = makes some angle θ with the positive ais, so we apply the c rotation R θ to T to get a new mapping [ [ [ T = R θ T, [ T cos θ sin θ a ã = [R θ [T = =, sin θ cos θ c 2

3 an et([ T ) = et([r θ [T ) = et([r θ ) et([t ) = et([t ). By the computation we i aove, Area( P ) = et([ T ). We also have that T sens the unit square S to a parallelogram P, an T sens S to a parallelogram P. These two parallelograms P an P iffer y a rotation, so they have the same area. Thus we see Area(P ) = Area( P ) = et([ T ) = et([t ). In particular, we have just proven that et([t ) precisely when T is invertile, ecause this is precisely when the image parallelogram P has nonzero area. Orientation an the sign of the eterminant: As we saw in the previous notes, there are actually two linear transformations which map the unit square S onto this parallelogram P, we can also have ([ ) [ T = In this case we see that ([, T ) [ a = et([t ) = a = Area(P ). [ a, [T = Why o we have the minus sign? To unerstan what s going on, it will help to lael the corners of the unit square S an the parallelogram P as in the picture elow.. iv i iii ii ii iii i iv What oes this laeling mean? The[ mapping T sens the vector e, which goes from i to ii in the square on the left to the vector. which also goes from i to ii in the parallelogram on the right. Similarly, the [ mapping T sens the vector e 2, which goes from i to iv in the square on a the left to the vector. which also goes from i to iv in the parallelogram on the right. Now, if we follow the laeling of the corners of the square in orer, as in i to ii to iii to iv, then we traverse along the ounary of the square counter-clockwise. However, if we follow the laeling of th ecorners of the parallelogram in orer, as in i to ii to iii to iv, we traverse along the ounary of the parallelogram clockwise. This means the mapping T reverse the irection we traverse along the ounary of the shape. In other wors, T reverse the orientation. We have iscovere the following general principle: et([t ) < T reverses orientation. This principle is eactly why we wrote et([t ) = Area(P ) efore. In general, if T : R 2 R 2 preserves orientation then et([t ) = Area(P ), ut if T reverses orientation then et([t ) = Area(P ). Higher imensions: So far we ve seen that the eterminant of a 2 2 matri is the area (up to a sign) of the parallelogram which is the image of the unit square. In fact, a similar thing 3

4 is true in higher imensions. Let [T e an n n matri, which we ve seen correspons to a linear map T : R n R n. Then T sens the unit cue S = { i : i =, 2,..., n} to a parallelpipe P, which is spanne y the columns of [T. Then et([t ) = Vol(P ), where Vol gives the n-imensional volume. We egin with a quick illustrative eample. Consier [T = a c e Then the image of the unit cue S uner T is where [ [ T a = c By slicing P with horizontal slices, we see, e >. P = {(, y, z) : (, y) P, z e},, S = {, y }, P = T ( S). Vol(P ) = e Area( P ) = e et([ T ). So, y any reasonale efinition of the eterminant for 3 3 matrices which fits with our efinition for 2 2 matrices, we must have et([t ) = e et([ T ) = e(a c). Eercise: Let [T e a 3 3 matri. Show that you can always perform a rotation to make the last colum of [T into. (Hint: what are the columns of [T?) e At this point, we can write own a reasonale formula for the eterminant of a 3 3 matri. Let [T = a e c f, g h i then ([ e et[t = c et g h ) ([ a f et g h ) ([ a + i et e Here we ve single out the last column, ut we can o the same thing y picking out any row or column. To o this properly, we nee some notation. Let [T = [A ij, so that the entry of [T in the ith row, jth column is A ij. Also, let [ T ij e the 2 2 matri you get from [T y crossing out the ith row an jth column. Then for any choice of j =, 2, 3 we can write ). et([t ) = ( ) +j A j et([ T j ) + ( ) 2+j A 2j et([ T 2j ) + ( ) 3+j A 3j et([ T 3j ), which computes et([t ) y eaning along the jth column. i =, 2, 3 we can write Alternatively, for any choice of et([t ) = ( ) i+ A i et([ T i ) + ( ) i+2 A i2 et([ T i2 ) + ( ) i+3 A i3 et([ T i3 ), which computes et([t ) y epaning along the ith row. The same iea will compute the eterminant of any square matri inuctively. That is, you write the eterminant of an n n matri as a sum of eterminants of (n ) (n ) matrices. We write the general formula as follows. Again, we let A ij e the entry of [T in the ith row, jth 4

5 column, an we let [ T ij e the (n ) (n ) matri you get from [T y crossing out the ith row an the jth column. The for any choice of j =, 2,..., n we compute et([t ) y epaning along the jth column using the formula et([t ) = ( ) +j A j et([ T j ) + ( ) 2+j A 2j et([ T 2j ) + + ( ) n+j A nj et([ T nj ). Alternatively, for any choice of i =, 2,..., n we compute et([t ) y epaning along the ith row using the formula et([t ) = ( ) i+ A i et([ T i ) + ( ) i+2 A i2 et([ T i2 ) + + ( ) i+n A in et([ T in ). We summarize some important properties of the eterminant here.. The eterminant is linear in each row an column. That is, if A is an n n matri an Ã is the same as A ecept that you multiply the ith row y c, then et(ã) = c et(a). Also, A an A 2 are the same ecept at the ith row an A is what you get y aing together the ith row of A an A 2 then et(a) = et(a ) + et(a 2 ). The same goes for columns. 2. Consequently, if A is an n n matri an c is a numer then et(ca) = c n et(a). 3. An n n matri A is invertile if an only if et(a). 4. In fact, et(a) is the n-imensional volume of the parallelpipe P which is the image of the unit cue S = {,..., n } uner the linear transformation associate to A. (You can prove this y inuction, in a very similar way we got the geometric interpretation for three imensions from the two-imensional version.) 5. Let A an B e n n matrices, then et(ab) = et(a) et(b). 6. Let A e an n n matri an let Ã e the matri you get y swapping ajacent two rows of A (or y swapping two ajacent columns). Then et(ã) = et(a) Trace: Another important numer associate to an n n matri is its trace. We let [T e an n n matri with A ij eing the entry in the ith row, jth column. Then the trace of [T is given y tr([t ) = A + A A nn, the sum of the entries of [T on the iagonal running from the top left of [T to its ottom right. Later on, we ll see that uner some conitions (for instance, if A ij = A ji ) that the trace measures an average istortion of T as a mapping. That is, if the trace is close to n then T oesn t istort istances too much, ut if the traces is very ifferent from n then it istorts istances a lot. Rememer that this guiline only hols if T is symmetric, that is if A ij = A ji. The trace satisfies the following properties:. If I is the n n ientity matri then tr(i) = n. 2. If A an B are square matrices an c is a numer then tr(a + B) = tr(a) + tr(b), tr(ca) = c tr(a), tr(ab) = tr(ba). Some special matrices: We close with some special types of square matrices. Let [T e a square matri, with entries A ij in the ith row, jth colum as aove. We say [T is iagonal if A ij = for i j. We say [T is upper triangular if A ij = for i > j an that [T is lower triangular if A ij = for i < j. Eercise: Why is a iagonal matri calle iagonal? How aout upper (or lower) triangular matrices? Eercise: Let [T e iagonal, an show that et([t ) = A A 22 A nn, tr([t ) = A + A A nn Eercise: Show that the same formula hols for a upper triangular an lower triangular matrices. 5

Rank, Trace, Determinant, Transpose an Inverse of a Matrix Let A be an n n square matrix: A = a11 a1 a1n a1 a an a n1 a n a nn nn where is the jth col

Rank, Trace, Determinant, Transpose an Inverse of a Matrix Let A be an n n square matrix: A = a11 a1 a1n a1 a an a n1 a n a nn nn where is the jth col Review of Linear Algebra { E18 Hanout Vectors an Their Inner Proucts Let X an Y be two vectors: an Their inner prouct is ene as X =[x1; ;x n ] T Y =[y1; ;y n ] T (X; Y ) = X T Y = x k y k k=1 where T an