1 2 2 Circulant Matrices

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Tracey Hancock
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1 Circulant Matrices General matrix a c d Ax x ax + cx x x + dx General circulant matrix a x ax + x a x x + ax. Evaluating the Eigenvalues Find eigenvalues and eigenvectors of general circulant matrix: a x x λx ax + x λ a x x λx x + ax λx ax + x λx x + ax add these λ (x + x )(a + )(x + x ) λ a + sutract these λ (x x )(a )(x x ) λ a Adifferent method: a x x λx λ a x x λx λ 0 x 0 λ 0 λ x 0 Ax λ Ix µ a a λ 0 0 λ Ax λ Ix 0 µ a (A λ I) x a a λ B a λ x x λ 0 0 λ x 0 the vector x is in the null suspace of the matrix B B does not exist determinant of B is zero

2 The determinant of the general matrix is: a c det ad c d a λ det (a λ) a λ 0 λ aλ + a 0 By convention, we select q λ ( a) ± ( a) 4(a ) 4a 4a +a ± +4 λ a ± λ a + λ a. Evaluating the Eigenvectors Now we can find expressions for the eigenvectors: a x x λ a x x a x ax + x a x x + ax x (a + ) x (a + ) x (a + ) x Equate expressions for first component : ax + x (a + ) x ax + x ax + x x x x x Equate expressions for second component : x + ax (a + ) x x + ax ax + x x x x x, SAME RESULT So the two components of the first eigenvector (corresponding to our convention for the first eigenvalue λ a + ) areequal: x x which thus lie along the identity line. x

3 x - - x - - The components of the first eigenvector of the circulant matrix lie along the line x x We often normalize the eigenvector to unit length, for reasons that will e more ovious later: ˆx + + corresponding to λ a + The components of the second eigenvector must simultaneously statisfy the two conditions: a x x x λ a x (a ) x x a x ax + x (a ) x a x x + ax (a +) x Equate expressions for first component : ax + x (a ) x ax + x ax x x x x x Equate expressions for second component : x + ax (a ) x x + ax ax x x x x x,sameresult One choice for the normalized second eigenvector is: ˆx + corresponding to λ a

4 Note that we could alternatively choose ˆx, ut we choose that + with the positively valued first component y convention. We can now plot the two normalized eigenvectors on the x-y plane: -D plane with normal (or canonical ) asis vectors shown in lack: + 0 and, and the two normalized eigenvectors shown in red: 0 " + # x " # + and x + + 4

5 . Projection of the Input Vector onto the Eigenvectors Clearly the two eigenvectors are orthogonal and normalized (unit length ˆx ˆx ), so they may e used as reference vectors to express any other -D + vector. For example, the vector x may e written in the normal + asis as: where the and are the respective projections of the vector onto the unit reference vectors. We can find the projections onto the two eigenvectors y evaluating the scalar product. For example, the project ion of the vector + x onto the first eigenvector ˆx + is: ˆx x + + " # + + ˆx x + " #

6 Projection of the vector onto the two eigenvectors of the circulant matrix. Note that the projection onto the second eigenvector ˆx is NEGATIVE, ecause the projection points in the opposite direction from the direction the vector points. This alternative representation of the input vector is equivalent and sometimes more useful (as we shall see): x x 0 where we just add a prime to the notation for the input vector x to denote 6

7 the representation in terms of the new asis vectors, that in turn were ased on the circulant matrix..4 Inverse of Circulant Matrix Recall the formula for the inverse of the general matrix: a c A A d c d ad c a As always, it is useful to confirm our memory: A d c a c A ad c a d ad c dc cd ad c a + a c + ad ad c 0 0 ad c 0 c + ad 0 which is the required identity matrix. You should confirm that the same result is otained from AA Now to the circulant case: a A A a A A a a a a a a a 0 0 a Sothistellsusrightawaywhatmustetrueaouta, to ensure that the inverse matrix exists. The determinant must not e zero: a det a a (a + )(a ) 6 0 a + 6 0AND a 6 0 a 6 ± Also note that the determinant of the circulant is the product of the eigenvalues: a det A det (a + )(a ) λ a λ which tells us that the matrix is invertile if neither eigenvalue is zero..4. Example: a a A a 7

8 First apply the formulas for the eigenvalues: λ a + + λ a 0 The two row vectors are identical, which tells us that the row suspace is -D: {x R } α + α α which includes all vectors along the diagonal, i.e., all vectors that are proportional to the first eigenvector of the general circulant matrix: Row suspace of the matrix A includes all vectors along the line x x. 8

9 Apply the matrix to a vector in the row suspace: A {x R } which confirms that vectors in the row suspace of this matrix are eigenvectors with eigenvalue λ +. The null suspace of this matrix includes all vectors that are orthogonal to the vectors in the row suspace: {x N } {x R } {x N } {x R } 0 u u {x N } α v v u + v 0 v u So the vectors in the null suspace are proportional to the normalized vector: {x N } + Check y applying the matrix to this vector: µ + 0 A {x N } 0 0 9

10 Null suspace of A includes all vectors along the line x x. The null suspace of this matrix is the lue dashed line and the row suspace is the red dotted line..4. Example: -a a A a + + λ a + +( ) 0 λ a ( ) 0

11 The two row vectors are identical ut for a change of sign: + + {x R } α + α α + + which tells us that the row suspace is -D. Note that the row suspace of A includes all of the vectors that were in the row suspace of A which means in turn that the null suspace of A must e identical to the row suspace of A : + {x N } + As always, it is advisale to check it y applying the matrix A to the formula for the nullspace vector: A {x N } includes all vectors along the diagonal, i.e., all vectors that are proportional to the first eigenvector of the general circulant matrix:

Solutions to Final Practice Problems Written by Victoria Kala Last updated 12/5/2015

Solutions to Final Practice Problems Written by Victoria Kala vtkala@math.ucsb.edu Last updated /5/05 Answers This page contains answers only. See the following pages for detailed solutions. (. (a x. See