$$

Diagonalizable or not

Problem

Let

$$A=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 2& 1& 0\\ 0& 0& 0& 2& 1\\ 0& 0& 0& 0& 2\end{array}\right].$$

For each distinct eigenvalue of

$A$, find its algebraic multiplicity and its geometric multiplicity. Then determine if

$A$ is diagonalizable or not.

Thought

For any

$n\times n$ matrix, its characteristic polynomial can be factored as

$$p_A(x)=(-1{)}^n(x-{\lambda }_1{)}^{m_1}(x-{\lambda }_2{)}^{m_2}\cdots (x-{\lambda }_q{)}^{m_q}.$$

Recall that for the eigenvalue

${\lambda }_i$,

${\mathrm{am}}_A({\lambda }_i)=m_i$ is the algebraic multiplicity and

${\mathrm{gm}}_A({\lambda }_i)=\mathrm{null}(A-{\lambda }_{i}I)$ is the geometric multiplicity. Intuitively,

${\mathrm{am}}_A({\lambda }_i)$ is the number of eigenvalues that equal to

${\lambda }_i$, while

${\mathrm{gm}}_A({\lambda }_i)$ is the number of independent eigenvectors that we can find.

It is known that

${\mathrm{gm}}_A({\lambda }_i)\le {\mathrm{am}}_A({\lambda }_i)$ for any eigenvalue

${\lambda }_i$. If

${\mathrm{gm}}_A({\lambda }_i)<{\mathrm{am}}_A({\lambda }_i)$ for some

${\lambda }_i$, then it means we cannot find enough independent eigenvectors for

${\lambda }_i$, indicating that

$A$ not diagonalizable.

Sample answer

By direct computation, the eigenvalues of

$A$ are

$\{1,1,2,2,2\}$. Thus, we have

${\mathrm{am}}_A(1)=2$ and

${\mathrm{am}}_A(2)=3$.

Again by direct computation, we have

$$A-I=\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 1& 1& 0\\ 0& 0& 0& 1& 1\\ 0& 0& 0& 0& 1\end{array}\right]\text{ and }A-2I=\left[\begin{array}{ccccc}-1& 0& 0& 0& 0\\ 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right].$$

Through row operations, we know

${\mathrm{gm}}_A(1)=\mathrm{null}(A-I)=2$ and

${\mathrm{gm}}_A(2)=\mathrm{null}(A-2I)=1$.

Since

$1={\mathrm{gm}}_A(2)<{\mathrm{am}}_A(2)=3$, the matrix

$A$ does not have enough independent eigenvectors for the eigenvalue

$2$, indicating that

$A$ is not diagonalizable.

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