Find an invertible matrix $Q \in M_3(\mathbb{R})$ such that $Q^{-1}AQ$ is a upper triangular matrix

Question

Let be \begin{align} A= \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 1 & 1 & 1 \end{bmatrix} \in M_3(\mathbb{R}) \end{align}

a) Find the minimal polynomial of $A$

b) Determine if $A$ is diagonizable over $\mathbb{R}$

c) Find an invertible matrix $Q \in M_3(\mathbb{R})$ such that $Q^{-1}AQ$ is a upper triangular matrix

My attempt:

a) First we find the characteristic polynomial \begin{align} det (A-tI)=(1-t)^{3} \end{align} Then, the minimal polynomial of $A$ is $(1-t)$

b) To verify if $A$ is diagonalizable, we have to verify if the dimension of the eigenspace associated to the eigenvalue 1, is 3. That is, to check if the algebraic multiplicity is equal to the geometric multiplicity of the eigenvalue 1.

\begin{align} \begin{pmatrix} 1-1 & 1 & 0\\ 0 & 1-1 & 0\\ 1 & 1 & 1-1 \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}&= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 1 & 1 & 0 \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}0\\ 0\\0\end{pmatrix} \\ \end{align}

\begin{align} \Longrightarrow \begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}0\\ 0\\0\end{pmatrix} \end{align}

$\Longrightarrow$ The dimension of the eigenspace associated to the eigenvalue 1, is zero. Then, $A$ is not diagonalizable.

My doubts:

• Am I correct in a) and b)?

• How can I find the matrix of c)? I would really appreciate your help

Answer 1

For $(a)$, it's easy to check by inspection that your minimal polynomial is not correct.

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For $(b)$, an eigenvector associated to the eigenvalue $1$ must satisfy $Av = v$, or $(A-I)v = 0$ as you put it. If $v = (x,y,z)$ then $(A-I)v = (y, 0, x+y)$, so $(A-I)v = 0$ implies $y = 0$ which then implies $x+y = x = 0$. Notice that this leaves $z$ free, and indeed it's easy to see $(0,0,1)$ is an eigenvector.

Of course, that still means $A$ is not diagonalizable.

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For $(c)$, if $A$ were diagonalizable, say $A = QDQ^{-1}$ with $D$ diagonal, then you could write $D = Q^{-1}AQ$ and since $D$ is diagonal it is also upper triangular.

Now, you've shown in $(b)$ that $A$ is not diagonalizable, but every matrix has a certain normal form that one could say is the next best thing. Do you think you can take it from here?

Answer 2

You got the right characteristic polynomial. But the dimension of the eigenspace associated with an eigenvalue is always greater than $0$. In this case, it is $1$: that eigenspace is spanned by $(0,0,1)$. So, let $e_1=(0,0,1)$. Now, you can check that a solution of $A.(x,y,z)=(x,y,z)+e_1$ is $(1,0,0)$. So, let $e_2=(1,0,0)$. And you can check that a solution of $A.(x,y,z)=(x,y,z)+e_2$ is $(-1,1,0)$. So, let$$Q=\begin{bmatrix}0&1&-1\\0&0&1\\1&0&0\end{bmatrix}$$(the columns of $Q$ are $e_1$, $e_2$, and $e_3$), and$$Q^{-1}.A.Q=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}.$$