0.6 Diagonalizability

A diagonal matrix is one whose elements not on the diagonal are equal to $0$ . The following matrix is one example. $$\begin{pmatrix}a & 0 & 0 & 0\\ 0 & b & 0 & 0\\ 0 & 0 & c & 0\\ 0 & 0 & 0 & d\\ \end{pmatrix}$$

A matrix $A$ is diagonalizable if there exists a matrix $V\in \mathbb{R}^{(n, n)}$ , $\det V\neq 0$ such that $VAV^{(-1)}=$ is diagonal. In such a case, the diagonal entries of $$ are the eigenvalues of $A$ .

Let's take an eigenvalue decomposition example to work backwards to this result.

Assume that the matrix $A$ has eigenvectors $v$ and $w$ and the respective eigenvalues ${}_v$ and ${}_w$ : $$Av={}_{v}v$$ $$Aw={}_{w}w$$

We can combine these two equations into an equation of matrices: $$A\begin{pmatrix}v & w\\ \end{pmatrix}=\begin{pmatrix}v & w\\ \end{pmatrix}\begin{pmatrix}{}_v & 0\\ 0 & {}_v\\ \end{pmatrix}$$

To simplify this equation, we can replace the eigenvector matrix with $V$ and the eigenvalue matrix with $$ . $$AV=V$$

Now, by multiplying both sides of the equation by $V^{(-1)}$ , we see the diagonalizability equation discussed above.

When is such a diagonalization possible? The condition is that the algebraic multiplicity equal the geometric multiplicity for each eigenvalue, ${}_i={}_i$ . This makes sense; basically, we are saying that there are as many eigenvectors as there are eigenvalues. If it were not like this, then the $V$ matrices would not be square, and therefore could not be inverted as is required by the diagonalizability equation . Remember that the eigenspace associated with a certain eigenvalue $$ is given by $\ker (A-I)$ .

This concept of diagonalizability will come in handy in different linear algebra manipulations later. We can however, see a time-saving application of it now. If the matrix $A$ is diagonalizable, and we know its eigenvalues ${}_i$ , then we can immediately find the eigenvalues of $A^2$ : $$A^2=VV^{(-1)}VV^{(-1)}=V^{2}V^{(-1)}$$

The eigenvalues of $A^2$ are simply the eigenvalues of $A$ , squared.