# Cayley-Hamilton theorem-eigenvalues

I am studying and doing exercises about the Cayley-Hamilton Theorem but I am having difficulties to understand this question:

I have the matrix $A$ as

\begin{bmatrix} 2&1\\ 1&2\\ \end{bmatrix}

I found the characteristic equation which is : $\lambda^2-4\lambda+3=0$ and eigenvalues of $\lambda=1$ and and $\lambda=3$

Then to apply Cayley-Hamilton I replaced the $\lambda$ in the characteristic equeation by A and obtained:

$A^2-4A+3=0$ and then I calculated matrix $A^2$ and verify that the Cayley-Hamilton is verified since,

$$A^2-4A+3I=\begin{bmatrix} 5&4\\ 4&5\\ \end{bmatrix} - 4\begin{bmatrix} 2&1\\ 1&2\\ \end{bmatrix} + 3\begin{bmatrix} 1&0\\ 0&1\\ \end{bmatrix} = \begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$ Considering the matrices $A,A^2,0$ the 2x2 zero matrix and $I$ the identity matrix and all linear combinations of them, so that they form a vector space.

How can I use the Cayley-Hamilton Theorem to give basis for this vector space?

I do not understand this part, should I found the eigenvectors for matrices $A,A^2,0$ and $I$?

Can anyone help on this I do not understand how Cayley-Hamilton can give a basis for this vector.

Thank you

## 1 Answer

Hint: The matrices $I$ and $A$ are linearly independent. However, the matrices $I,A,A^2$ are not linearly independent.

• So because they are linearly independent ( they only have the trivial solution x=y=0) the basis are the vector (2,1),(1,2)(1,0) and (0,1) ? Feb 9, 2017 at 1:34
• What? No. Our vector space consists of square matrices, not column vectors. The basis should consists of square matrices, not column-vectors. One possible basis is $\{I,A\}$. Feb 9, 2017 at 2:02
• So I was checking and did: -8x-4y=0,-4x-8y=0,3x+0y=0,0x+y=0 corresponding to matrix A and I multiplied by the coefficients -4 and 3 ( A^2-4A+12I) with solutions x=y=0 (linearly independent) and then 5x+4y=0,4x+5y=0,-8x-4y=0,-4x-8y=0,3x+0y=0,0x+y=0 with no trivial solutions so they are linearly dependent. Am I doing something wrong? can you help me on this? thanks Feb 9, 2017 at 2:02
• Sorry, I understood now. Thank you for your help Feb 9, 2017 at 2:03
• Hi, sorry to ask again, but just to clarify: you are saying that {A,I} is a possible basis because $\left(\begin{array}{cc|cccc} 2 & 1 & 0\\ 1 & 0 & 0 \\ 1 & 0 & 0\\ 2 & 1 & 0\\ \end{array}\right)$ x=y=0 Lin.Indp so they form a basis. However {I,A,$A^2$}= $\left(\begin{array}{ccc|cccc} 5 & 2 & 1 & 0\\ 4 & 1 & 0& 0 \\ 4 & 2 & 0 & 0\\ 5 & 1 & 1 & 0\\ \end{array}\right)$ no trivial solution no lin.indp then don't form a basis. Mar 3, 2017 at 13:56