How do I show that a given lambda value is an eigen value of a matrix? Also how do I show that a matrix is not diagonalizable based on my calculations from the equation $\det(A-\lambda I)=0$

For the first part say the matrix is:

$$
        \begin{pmatrix}
        1 &amp; -2 &amp; 3 \\
        0 &amp; -1 &amp; 3 \\
        -1 &amp; 2 &amp; -2 \\
        \end{pmatrix}
$$
is $\lambda=1$ an eigenvalue for the matrix?

Also I wouldn't mind finding out the code for lambda on here and the code for putting a set vectors in brackets as related to span.

Thanks!
To find the eigenvalues of $A=\begin{bmatrix} 1&amp;-2&amp;3\\0&amp;-1&amp;3\\-1&amp;2&amp;-2\end{bmatrix}$ we compute the roots of the characteristic polynomial, $p_A(t)=det(tI-A)$. Using Laplace Expansion we have:
$$p_A(t)=det\begin{bmatrix} t-1&amp;2&amp;-3\\0&amp;t+1&amp;-3\\1&amp;-2&amp;t+2\end{bmatrix}=(t-1)det\begin{bmatrix} t+1&amp;-3\\-2&amp;t+2 \end{bmatrix}+det\begin{bmatrix} 2&amp;-3\\t+1&amp;-3 \end{bmatrix}=(t-1)[(t+1)(t+2)-6]+[-6+3t+3]=(t-1)(t^2+3t-1)$$
From factored form, we see that the eigenvalues of $A$ are 
$\lambda_1=1, \lambda_2=\frac{-3+\sqrt {13}}{2}, \lambda_3=\frac{-3-\sqrt {13}}{2}$. So to answer the first question: yes, the matrix has $1$ as an eigenvalue. Now, because the matrix is $3\times 3$ and it has $3$ distinct eigenvalues, it is diagonalizable.

Note that showing an $n\times n$ matrix is diagonalizable is equivalent to showing that the algebraic multiplicities of its eigenvalues equals their geometric multiplicities. In the special case where it has $n$ distinct eigenvalues, this condition is immediately seen to hold true, and the matrix is diagonalizable.
After subtracting the eigenvalues $\lambda$ from diagonal elements are some rows or columns or after some linear combinations the same? Does it vanish?

$$
        \begin{pmatrix}
        0 &amp; -2 &amp; 3 \\
        0 &amp; -2 &amp; 3 \\
        -1 &amp; 2 &amp; -3 \\
        \end{pmatrix}
$$