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HomeMatlabWhat Is a Diagonalizable Matrix? – Nick Higham

What Is a Diagonalizable Matrix? – Nick Higham


A matrix A inmathbb{C}^{ntimes n} is diagonalizable if there exists a nonsingular matrix Xinmathbb{C}^{ntimes n} such that X^{-1}AX is diagonal. In different phrases, a diagonalizable matrix is one that’s much like a diagonal matrix.

The situation X^{-1}AX = D = mathrm{diag}(lambda_i) is equal to AX = XD with X nonsingular, that’s, Ax_i = lambda_ix_i, i=1colon n, the place X = [x_1,x_2,dots, x_n]. Therefore A is diagonalizable if and provided that it has an entire set of linearly impartial eigenvectors.

A Hermitian matrix is diagonalizable as a result of the eigenvectors could be taken to be mutually orthogonal. The identical is true for a traditional matrix (one for which A^*A = AA^*). A matrix with distinct eigenvalues can be diagonalizable.

Theorem 1.

If Ainmathbb{C}^{ntimes n} has distinct eigenvalues then it’s diagonalizable.

Proof. Let A have eigenvalues lambda_1,lambda_2,dots,lambda_n with corresponding eigenvectors x_1,x_2,dots,x_n. Suppose that y = sum_{i=1}^n alpha_i x_i = 0 for some alpha_1,alpha_2,dots,alpha_n. Then

notag begin{aligned}  0 &= (A-lambda_2 I)cdots (A-lambda_n I)y    = sum_{i=1}^n alpha_i (A-lambda_2 I)cdots (A-lambda_n I) x_i    &= sum_{i=1}^n alpha_i (lambda_i-lambda_2)cdots(lambda_i-lambda_n)x_i    =  alpha_1 (lambda_1-lambda_2)cdots(lambda_1-lambda_n)x_1, end{aligned}

which suggests alpha_1 = 0 since lambda_1ne lambda_j for jge 2 and x_1ne0. Premultiplying y = sum_{i=2}^n alpha_i x_i = 0 by prod_{j=3}^n(A-lambda_j I) exhibits, in the identical means, that alpha_2 = 0. Persevering with on this means we discover that alpha_1 = alpha_2 = cdots = alpha_n = 0. Subsequently the x_i are linearly impartial and therefore A is diagonalizable.

A matrix can have repeated eigenvalues and be diagonalizable, as diagonal matrices with repeated diagonal entries present. What is required for diagonalizability is that each k-times repeated eigenvalue has k linearly impartial eigenvectors related to it. Equivalently, the algebraic and geometric multiplicities of each eigenvalue should be equal, that’s, the eigenvalues should all be semisimple. One other equal situation is that the diploma of the minimal polynomial is the same as the variety of distinct eigenvalues.

The best instance of a matrix that isn’t diagonalizable is bigl[begin{smallmatrix} 0 & 1  0 & 0 end{smallmatrix}bigr]. This matrix is a 2times 2 Jordan block with the eigenvalue 0. Diagonalizability is definitely understood by way of the Jordan canonical type: A is diagonalizable if and provided that all of the Jordan blocks in its Jordan type are 1times 1.

Most matrices are diagonalizable, within the sense that the diagonalizable matrices are dense in mathbb{C}^{ntimes n}, that’s, any matrix in mathbb{C}^{ntimes n} is arbitrarily near a diagonalizable matrix. This property is beneficial as a result of it may be handy to show a outcome by first proving it for diagonalizable matrices after which arguing that by continuity the outcome holds for a common matrix.

Is a rank-1 matrix A = xy^* diagonalizable, the place x,yinmathbb{C}^{ntimes n} are nonzero? There are n-1 zero eigenvalues with eigenvectors any set of linearly impartial vectors orthogonal to y. If y^*x ne 0 then y^*x is the remaining eigenvalue, with eigenvector x, which is linearly impartial of the eigenvectors for 0, and A is diagonalizable. If y^*x = 0 then all of the eigenvalues of A are zero and so A can’t be diagonalizable, as the one diagonalizable matrix whose eigenvalues are all zero is the zero matrix. For the matrix bigl[begin{smallmatrix} 0 & 1  0 & 0 end{smallmatrix}bigr] talked about above, x = bigl[{1 atop 0} bigr] and y = bigl[{0 atop 1} bigr], so y^*x = 0, confirming that this matrix just isn’t diagonalizable.

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