# For which $t \in \mathbb{R}$ is matrix diagonalizable

$\begin{bmatrix} 1 & t & 25 \\ 0 & t & t+1 \\ 0 & 0 & -1 \end{bmatrix}$

I calculated the characteristic polynomial which is $p(\lambda) = (\lambda^2 - 1)(t-\lambda)$. But how do I find specific $t$, for which the matrix is not diagonalizable?

• The only values you need to test are those that give the polynomial a repeated root. – Gerry Myerson Jun 1 '13 at 11:17
• – rschwieb Jun 1 '13 at 12:06

Hint: If there are $3$ distinct eigenvalues, then the matrix is diagonalizable. Hence there are only two possible values for which the matrix may not be diagonalizable.

If $t\neq \pm 1$ then characteristic polynomial has $3$ simple roots so the given matrix $A$ is diagonalizable

If $t=1$ then $1$ is an eigenvalue with multiplicity $2$ and we find $\ker(A-I_3)=\mathrm{span}((1,0,0)^T)$ then the matrix isn't diagonalizable since $$\dim\ker(A-I_3)=1\neq 2$$

If $t=-1$ then $-1$ is an eigenvalue with multiplicity $2$ and we find $\ker(A+I_3)=\mathrm{span}((\frac{1}{2},1,0)^T),(-\frac{25}{2},0,1)^T)$ then the matrix is diagonalizable since $$\dim\ker(A+I_3)=2$$

The answers above address the case when all the eigenvalues are distinct and thus their respective eigenvectors are linearly independent (and then the matrix is diagonalizable).

Yet in your case, also when $\,t=-1\,$ is an eigenvalue the matrix is diagonalizable, since as you can readily check we get the eigenspace corresponding to this eigenvalue is

$$V_{-1}=\{(x,y,z)\in\Bbb R^3\;;\;2x-y+25=0\}\implies \dim V_{-1}=2$$

So you can find two linearly independent eigenvectors corresponding to $\,\lambda=-1\,$ and one linearly independent of these two corr. to $\,\lambda =1\,$ and the matrix is diagonalizable.

Not so when $\,t=1\,$ (why? Check the corr. eigenspace...)

If $p$ has distinctive eigenvalues, then the matrix is diagonalizable, so it suffices to examine the case $t=\pm 1$.

Write our matrix as $A(t)$.
We know that a matrix is diagonalizable if and only if the geometric and the algebraic multiplicities of eigen-values are all the same. Now as geometric multiplicities are always less than or equal to the algebraic ones, if the characteristic polynomial has no repeated factors, then the matrix is definitely diagonalizable. In our case, this happens when $t\not=1,-1$.
When $t=1$, then $A(1)-I$ has rank $2$, so the geometric multiplicity of the eigen-value $1$ is 1, showing that $A(1)$ is not diagonalizable.
If $t=-1$, then $A(-1)+I$ has rank $=1$, so the geometric multilpicity is equal to the algebraic one, which $=2$. Hence $A(-1)$ is diagonalizable.
Inform me of any errors, or inappropriate points.