Is the matrix diagonalizable?

Is $A = \left( {\matrix{ 3 & 0 \cr 1 & 3 \cr } } \right)$ diagonalizable? Prove it using characteristic/minimal polynomial

$$f_A(x)=(x-3)^2$$

I think the minimal polynomial is $m_A = x-3 \ne f_A(x)$ and therefore, $A$ isn't diagonalizable. Am I right?

• No, the minimal polynomial is not $x-3$, but $(x-3)^2$. The fact that $m_A\ne f_A$ doesn't imply nondiagonalizability anyhow. Dec 1, 2014 at 21:55
• @egreg, we haven't learned yet about algebric/geometric multiplicity. What should I do then? Dec 1, 2014 at 21:58

The polynomial $x-3$ is not the minimal polynomial, since $$A-3I_2=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$$ which is not the zero matrix.

The minimal polynomial is indeed $(x-3)^2$ as you can easily verify. Since this polynomial doesn't have pairwise distinct roots, the matrix is not diagonalizable.

The fact that $m_A\ne f_A$ doesn't imply the matrix is non diagonalizable: think to the identity matrix $I_2$, which has minimal polynomial $x-1$ and characteristic polynomial $(x-1)^2$, but is certainly diagonalizable.

Another way to see $A$ is not diagonalizable is by computing the geometric multiplicity of the eigenvalue $3$, which is $1$, while its algebraic multiplicity is $2$.

• So if the characteristic polynomial has $(x-\lambda)^k$ where $k>1$ then $A$ isn't diagonalizable? Dec 1, 2014 at 22:02
• @AlonAlon Yes, that's it. But this is usually proved with Jordan canonical form; it's not the only route, though. Dec 1, 2014 at 22:04

that $A = \pmatrix{3 & 0\cr 1 & 3}$ can be show without invoking minimal polynomial of this matrix. we will use the following two facts:

(a) $A$ is diagonalizable if and only if $A - 3I,$

(b) eigenvalues of a lower triangular matrix are its diagonal elements,

here is the argument that $A$is not diagonalizable. suppose it is diagonalizable. then so is $A - 3I,$ but the eigenvalues of $A - 3I$ are $0$ twice. this implies $A-3I = \pmatrix{0 & 0 \cr 1 & 0} = S \pmatrix{0 & 0 \cr 0 & 0} S^{-1} = \pmatrix{0 & 0 \cr 0 & 0}.$ this is a contradiction, therefore neither $A$ nor $A-3I$ is diagonalizable.