We saw in the previous lecture that $A \in M_{n \times n}$ is diagonalizable if there is a basis $\beta = \{v_1, v_2,...,v_n\}$ of $\mathbf{R}^n$ such that

$$ \begin{align*} [L_A]_{\beta}^{\beta} = \begin{pmatrix} \lambda_1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \lambda_n \end{pmatrix} \end{align*} $$

Where each vector of this matrix is $Av_j = \lambda_j v_j$. $v_j$ is an eigenvector and $\lambda_j$ is called an eigenvalue.
Let the standard basis of $\mathbf{R}^n$ be $\alpha$, then we know that $[L_A]_{\alpha}^{\alpha} = A$. We also know that we can find a change of coordinates matrix such that

$$ \begin{align*} [L_A]_{\beta}^{\beta} &= [I_{\mathbf{R}^n}]_{\alpha}^{\beta}[L_A]_{\alpha}^{\alpha}[I_{\mathbf{R}^n}]_{\beta}^{\alpha} \\ &= Q^{-1}AQ. \end{align*} $$

So now to diagonalize a matrix, we need to find these eigenvectors and eigenvalues that satisfy $Av_j = \lambda_jv_j$ for $j = 1,...,n$ and $v \neq \bar{0}$.
Finding such a basis is equivalent to saying that $A$ is diagonalizable if and only if there is a basis $\beta$ consisting of eigenvectors.
In fact, to diagonalize $A$, we need $n$ linearly independent eigenvectors of $A$.
Given an eigenvalue, it is easy to find the eigenvectors. $E_{\lambda} = N(A - \lambda I_n) = \{ \text{ eigenvectors of $A$ with eigenvalue $\lambda$} \}\cup \{\bar{0}\} $
Given $E_{\lambda} = N(A - \lambda I_n)$, we can find a basis of independent eigenvectors. The number of eigenvectors is the dimension of $E_{\lambda}$.
This is all good but it required us to know the eigenvalues. To find the eigenvalues, we discovered last time that the eigenvalues of $A$ are the real roots of the characteristic polynomial $f(t) = \det(A - tI_n)$.

This is great and gives up the following plan to diagonalize $A$:

Find all the eigenvalues of $A$, $\lambda_1,...,\lambda_k$.
Find a basis for $E_{\lambda_j}$.
Determine if these bases combine to form a basis $\beta$ for $\mathbf{R}^n$
Compute $Q, Q^{-1}$ etc.

Example

Diagonalize $\begin{align*} A = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 2 & 0 \\ -2 & 0 & 3 \end{pmatrix} \end{align*}$ If possible.

Following the above plan:

Find all the eigenvalues. To do this we need to find the roots of the polynomial $f(t)$
$$ \begin{align*} \det (A - tI_3) &= \det \begin{pmatrix} -t & 0 & 1 \\ 0 & 2-t & 0 \\ -2 & 0 & 3-t \end{pmatrix} \\ &= (-1)^{2+2} \det \begin{pmatrix} -t & 1 \\ -2 & 3-t \end{pmatrix} \\ &= (2-t)((-t)(3-t) + 2) \\ &= (2-t)(t^2 -3t + 2) \\ &= (2-t)(t-1)(t-2) \\ &= -(t-2)^2(t-1) \end{align*} $$
The solutions are $\lambda_1 = 1$ and $\lambda_2 = 2$.
We know from the example of the previous lecture that for $\lambda_1 = 1$, the eigenspace consists of the vectors $E_{\lambda_1} = span\{ (1,0,1)\}$. For $\lambda_2 = 2$, we need again to put $A - 2I_3$ in row echelon form to find the eigenvectors.
$$ \begin{align*} A - 2I_3 = \begin{pmatrix} -2 & 0 & 1 \\ 0 & 0 & 0 \\ -2 & 0 & 1 \end{pmatrix} &R_3 \rightarrow -R_1 + R_3 \begin{pmatrix} -2 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\ \begin{pmatrix} -2 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} &R_1 \rightarrow -\frac{1}{2}R_1 \begin{pmatrix} 1 & 0 & -\frac{1}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align*} $$
Therefore,
$$ \begin{align*} N(A - 2I_3) &= \{(x_1, x_2, x_3) \ | \ x_1 = \frac{1}{2}x_3\} \\ &= \{(\frac{1}{2}t_2, t_1, t_2) \ | \ t_1, t_2 \in \mathbf{R}\} \\ &= span\{(0,1,0), (\frac{1}{2},0,1)\} \end{align*} $$
Next, we need to determine if $\{(1,0,1),(0,1,0),(\frac{1}{2},0,1)\}$ is a basis for $\mathbf{R}^3$. This means we want to know if they are linearly independent. We can do by seeing if
$$ \begin{align*} a(1,0,1) + b(0,1,0) + c(\frac{1}{2}, 0,1) = 0 \end{align*} $$
has only the zero vector as the solution. This translate to the following system of equations.
$$ \begin{align*} a + \frac{1}{2}c &= 0 \\ b &= 0 \\ a + \frac{1}{2}c &= 0 \end{align*} $$
We can put this in a matrix and reduce it to row echelon form.
$$ \begin{align*} \begin{pmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} &R_3 \rightarrow -R_1 + R_3 \begin{pmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{pmatrix} \end{align*} $$
From this we see that the vectors are linearly independent and $\beta$ is therefore a basis.
The last step is to compute the change of coordinate matrix $Q = [I]_{\beta}^{\alpha}$. This matrix is easy to compute. Taking each vector from the basis $\beta$ and expressing it as a linear combination of the standard basis $\alpha$ will just give us the same vector back. Therefore,
$$ \begin{align*} Q = \begin{pmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \end{align*} $$
We also need to compute $Q^{-1}$ but we can do saw using the usual method (row reduced echelon form with the identity matrix on the right).
$$ \begin{align*} Q^{-1} = \begin{pmatrix} 2 & 0 & -1 \\ 0 & 1 & 0 \\ -2 & 0 & 2 \end{pmatrix} \end{align*} $$

Therefore, we finally have

$$ \begin{align*} B = [L_A]_{\beta}^{\beta} &= [I_{\mathbf{R}^n}]_{\alpha}^{\beta}[L_A]_{\alpha}^{\alpha}[I_{\mathbf{R}^n}]_{\beta}^{\alpha} \\ &= Q^{-1}AQ \\ &= \begin{pmatrix} 2 & 0 & -1 \\ 0 & 1 & 0 \\ -2 & 0 & 2 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 2 & 0 \\ -2 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\\ &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \end{align*} $$

Remarks:
Remark 1: We knew $B$ already. We knew what it would look like since the diagonal entries should be the eigenvalues.

Remark 2: We can solve for $A$ to get this really nice factorization.

$$ \begin{align*} A^k &= Q \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^k & 0 \\ 0 & 0 & 2^k \end{pmatrix} Q^{-1} \end{align*} $$

Eigenvectors of distinct eigenvalues are Linearly Independent

Question: is the strategy of combining bases of various eigenspaces $E_{\lambda}$’s to form larger linearly independent subsets always work or useful?

The answer is yes and is demonstrated by the next lemma,

Lemma

Suppose $\lambda_1, \lambda_2$ are distinct eigenvalues of $A$. If $w \neq \bar{0}$ is in $E_{\lambda_2}$, then it is not in $E_{\lambda_1}$.

Proof:

Let $\beta = \{v_1,...,v_k\}$ be a basis of $E_{\lambda_1}$ consisting of eigenvectors. Suppose for the sake of contradiction that $w \neq 0 \in E_{\lambda_1}$ for some $w\in E_{\lambda_2}$. Since $w$ is in the eigenspace of $\lambda_1$, this means that $w$ is in the span of of the basis $\beta$ by the definition of a basis. Therefore, we can write $w$ can be written as a linear combination of the vectors in $\beta$ for some unique scalars $a_1, a_2, ...,a_k$. (unique because it’s a basis)

$$ \begin{align*} w &= a_1v_1 + ... + a_kv_k \\ Aw &= a_1Av_1 + ... + a_kAv_k \\ \lambda_2w &= a_1\lambda_1 v_1 + ... + a_k\lambda_1 v_k \text{ (because $Av = \lambda v$ for eigenvectors)} \\ w &= \frac{\lambda_1}{\lambda_2} a_1v_1 + ... + \frac{\lambda_1}{\lambda_2} a_k v_k \\ \end{align*} $$

But since $a_1,...,a_k$ are unique scalars then we must have $\lambda_1 = \lambda_2$. But this is a contradiction since $\lambda_1 \neq \lambda_2. \ \blacksquare$

This lemma implies that the eigenvectors corresponding to distinct eigenvalues always form linearly independent sets when combined. The following corollary states exactly this.

Corollary

If $\lambda_1 \neq \lambda_2$ and $\beta_1, \beta_2$ are basis for $E_{\lambda_1}$ and $E_{\lambda_2}$ respectively, then $\beta_1 \cup \beta_2$ is linearly independent.

This means that we didn’t need to do step 3 from our algorithm to check that our eigenvectors are linearly independent. They will be! The proof for the corollary is a homework question!

Is Every Matrix Diagonalizable?

The answer is no! For some matrices, we won’t be able to find $n$ linearly independent eigenvectors. One test that we did last time is to check the characteristic polynomial’s roots or in other words, solve $\det(A - tI_2) = 0$.

$$ \begin{align*} \det(A - tI_2) = \det \begin{pmatrix} -t & -1 \\ 1 & -t \\ \end{pmatrix} = t^2 + 1 \end{align*} $$

This polynomial has no real roots and so we don’t have eigenvalues. Is the only test? No, we will develop a better test for this.

References

Math416 by Ely Kerman