We saw in the previous lecture that $A \in M_{n \times n}$ is diagonalizable if $A$ has $n$ linearly independent eigenvectors $\beta = \{v_1, v_2,...,v_n\}$. In that case,

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

$v_j$ is an eigenvector and $\lambda_j$ is called an eigenvalue. We developed a plan to find these and eigenvalues and eigenvectors.

Find these eigenvalues $\lambda_1, ..., \lambda_2$
Find a basis for $\beta_i$ for each eigenspace $E_{\lambda_i}$
If $\sum_{j=1}^{k}\dim(E_{\lambda_i}) = n$, then collect all the separate bases and that's our basis $\beta = \beta_1 \cup ... \cup \beta_k$. This step works because we proved in the last lecture the corollary $\lambda_1 \neq \lambda_2 \rightarrow \beta_1 \cup \beta_2$ is linearly independent.

Examples of Non-Diagonalizable Matrices

Of course the plan above might not work and there are different ways, a matrix could fail to be diagonalized.

$$ \begin{align*} A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{align*} $$

This matrix has no eigenvalues so it can’t be diagonalized.

$$ \begin{align*} B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{align*} $$

This matrix has only one eigen value.

$$ \begin{align*} det(A - tI_2) = \begin{pmatrix} 1-t & 1 \\ 0 & 1-t \end{pmatrix} = (1-t)^2 = 0 \rightarrow t = 1 \end{align*} $$

This implies

$$ \begin{align*} E_{\lambda_1} = N(B - 1I_n) = N \left( \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \right) = \{ (t, 0) \ | \ t \in \mathbf{R} \} \end{align*} $$

So we don’t have enough linearly independent vectors in $\beta$.

Is There a Better Diagonlization Test?

Today, we will refine our answer to the question “Is $A$ diagonalizable?”

Definition

A polynomial $f(t)$ splits over $\mathbf{R}$ if there are scalars $c,a_1,...,a_k \in \mathbf{R}$ such that $$ \begin{align*} f(t) = c(t-a_1)(t - a_2)...(t-a_k) \end{align*} $$

In other words, can completely factorize the polynomial?

Example 1: $t^2 + 1$ doesn’t split over $\mathbf{R}$. It does however split over $\mathbf{C} as (t - i)(t + i)$.

Example 2: $t^2 - 2t + 1 = (t - 1)(t - 1)$ splits over $\mathbf{R}$.

So now what does splitting have to do with diagonalizability? The following theorem explains this

Theorem 1

If $A$ is diagonalizable, then its characteristic polynomial splits over $\mathbf{R}$

Note that the the converse is false. If the characteristic polynomial splits over $\mathbf{R}$, it doesn’t necessarily means that $A$ is diagonalizable. (See example 2 above)

Eigenvalues of Similar Matrices

Before going into the proof of theorem 1 above, the following proposition will be useful

Proposition

If $B = Q^{-1}AQ$, then $$ \begin{align*} \det(B - tI_n) = \det(A - tI_n) \end{align*} $$

This means that if $A$ and $B$ are similar, then they have the same eigenvalues.

Proof:

Observe that

$$ \begin{align*} 1 = \det(I_n) = \det(QQ^{-1}) = \det(Q)\det(Q^{-1}). \end{align*} $$

Therefore,

$$ \begin{align*} \det(B - tI_n) &= \det(Q^{-1}AQ - tI_n) \\ &= \det(Q^{-1}AQ - QtI_nQ^{-1}) \\ &= \det(Q (A - tI_n)Q^{-1}) \text{ (factor out Q on the left ..) }\\ &= \det(Q) \det((A - tI_n)) \det(Q^{-1}) \\ &= \det(Q) \det(Q^{-1}) \det((A - tI_n)) \text{ (they are just real numbers)}\\ &= \det((A - tI_n)). \ \blacksquare \\ \end{align*} $$

Note here that in step 2, $QtI_nQ^{-1} = tQQ^{-1} = tQtQ^{-1} = tI_n$

Proof of Theorem 1

Suppose that $A$ is diagonalizable. This means that there exists a basis $\beta$ such that

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

If we let $\alpha$ be the standard basis, then know that there exists a change of coordinate matrix $Q = [I]^{\alpha}_{\beta}$ such that

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

This means that $[L_A]_{\beta}^{\beta}$ and $A$ are similar matrices. But by the previous proposition, this means that we have

$$ \begin{align*} \det ([L_A]_{\beta}^{\beta} -tI_n) &= \det(A - tI_n). \end{align*} $$

On the other hand,

$$ \begin{align*} \det ([L_A]_{\beta}^{\beta} -tI_n) &= \begin{pmatrix} \lambda_1 -t & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \lambda_n -t \end{pmatrix} \\ &= (\lambda_1 - t)...(\lambda_n - t). \ \blacksquare \end{align*} $$

Algebraic and Geometric Multiplicities of Eigenvalues

There is another condition for diagonalizability but we need a few more definitions.

Definition

The algebraic multiplicity of $\lambda$ is the number of times $\lambda - t$ divides $\det(A - tI_n)$.

An an example if $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ then, $\lambda = 1$ has algebraic multiplicity 2 because $\det(A - I_n) = (1-t)^2$.

Definition

The geometric multiplicity of $\lambda$ is $\dim(E_{\lambda})$.

For the same example above. The geometric multiplicity of $\lambda = 1$ is 1 because the nullspace is spanned by one vector.

Given these definitions we can now introduce the following theorem.

Theorem 2

Geometric multiplicity of $\lambda \leq$ the algebraic multiplicity of $\lambda$.

Proof
Let $\lambda$ be an eigenvalue of $A \in M_{n \times n}$ and let the geometric multiplicity of $\lambda$ be $k$. Now the goal is to relate the characteristic polynomial of $\lambda$ to the dimension of $E_{\lambda}$.

Let $\{v_1,...,v_k\}$ be a basis of $E_{\lambda}$. We know that $k \leq n$ so extend this basis to a basis for $\mathbf{R}^n$.

Theorem

$A$ is diagonalizable if and only if

$\det(A - tI_n)$ splits over $\mathbf{R}$
For each eigenvalue $\lambda$, geometric multiplicity = algebraic multiplictiy

References

Video Lectures from Math416 by Ely Kerman.