I have been given two matrices that are defined as follows:
$$A = \left(\begin{array}{rrrr}
-1&2&-3&-1\\%
8&-7&12&4\\%
6&-6&10&3\\%
2&-2&3&2\\%
\end{array}\right), B=
\left(\begin{array}{rrrr}
-1&-4&7&-6\\%
0&11&-25&25\\%
-4&8&-25&28\\%
-4&4&-16&19\\%
\end{array}\right) \in \mathbb{Q}^{4 \times 4}.$$
I need two prove or disprove that the matrices A and B are similar.
The matrices A and B have the same eigenvalue, which are equal to $1$.
Furthermore, I found out that the eigenspaces of the different matrices are not equal. The eigenspace for a given matrix A and eigenvalue a is defined as the solution set of:
$$V_a(A) = \mathbb{L}(A – a \cdot E_n, 0)$$
The geometric multiplicity for a matrix $A$ of an eigenvalue $a$ is defined as the dimension of the eigenspace:
$$g_a(A) = dim (\mathbb{L}(A – a \cdot E_n, 0))$$
For the eigenspaces and geometric multiplicity of the matrices A and B, it follows:
$$V_1(A) = \mathbb{L}(A – 1 \cdot E_4, 0) = \{x_2 \cdot \left(\begin{array}{rrr}
1\\%
1\\%
0\\%
0\\%
\end{array}\right)
+ x_3 \cdot \left(\begin{array}{rrr}
-3/2\\%
0\\%
1\\%
0\\%
\end{array}\right) + x_4 \cdot \left(\begin{array}{rrr}
-1/2\\%
0\\%
0\\%
1\\%
\end{array}\right) | x_2, x_3, x_4 \in \mathbb{Q}\}$$
$$\Rightarrow dim (\mathbb{L}(A – 1 \cdot E_4, 0)) = 3$$
$$\Rightarrow g_1(A) = 3$$
$$V_1(B) = \mathbb{L}(B – 1 \cdot E_4, 0) = \{x_3 \cdot \left(\begin{array}{rrr}
-3/2\\%
5/2\\%
1\\%
0\\%
\end{array}\right) + x_4 \cdot \left(\begin{array}{rrr}
2\\%
-5/2\\%
0\\%
1\\%
\end{array}\right) | x_3, x_4 \in \mathbb{Q}\}$$
$$\Rightarrow dim (\mathbb{L}(B – 1 \cdot E_4, 0)) = 2$$
$$\Rightarrow g_1(B) = 2$$
Because the geometric multiplicity of both matrices are not equal I would assume that matrices A and B are not similar. But how do I prove this formal.
I know if two matrices (e.g. A and B) are similar, it follows:
$$A = T^{-1} \cdot B \cdot T \, \text{for} \, T \in GL_n(\mathbb{Q})$$
How would I use this definition to disprove the given statement.
Best Answer
If $A$ and $B$ are similar matrices, then for every eigenvalue $a$, the eigenspaces $V_a(A)$ and $V_b(B)$ are isomorphic (and hence have the same dimension). In your notation, if $A = T^{-1}BT$, then $v$ is an eigenvector of $A$ if and only if $T\cdot v$ is an eigenvector of $B$ (verify this). So, the isomorphism is given by $v \mapsto T\cdot v$, from $V_a(A)$ onto $V_a(B)$.
However, you showed that these spaces have different dimensions, hence $A$ and $B$ are not similar. (this is just the contrapositive of the statement above).
Alternatively, if you know about Jordan Canonical form, you can answer this question as follows: $\dim V_a(A) = 3$ means the JCF of $A$ has $3$ Jordan blocks corresponding to the eigenvalue of $1$. However, $\dim V_a(B) = 2$ means that in the JCF of $B$ there are only $2$ Jordan Blocks corresponding to the eigenvalue $1$.
Since similar matrices (whose characteristic polynomials split) have the same JCF (up to permutation of the blocks), it follows that $A$ and $B$ are not similar.
By the way, in your particular example, the Jordan canonical forms are: \begin{align} J_A = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \quad \text{and} \quad J_B = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{align} respectively (which are not similar)