I'm currently doing reading into calculating the eigenvalues and eigenvectors of a $3\times3$ matrix and come across the following confusion:
In one of the resources that I'm using, it gives the identity matrix has this:
$$I = \begin{bmatrix} \lambda& 0 & 0 \\ 0 & \lambda & 0\\ 0& 0& \lambda \end{bmatrix}$$
But in most cases, and, in the original definition the identity matrix has $1$'s down the diagonal line where everything else is $0$'s.
I know that $\lambda$ is a scalar value, and therefore would it be correct to assume that: $\lambda = 1$?
Therefore, suppose that I have this matrix:
$$A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & -1 & 3\\ 4& 5& 10 \end{bmatrix}$$
Can I therefore subtract each of the elements in $A$ by the expression that $\lambda = 1$? If this is wrong, how am I missing the point of what the $\lambda$ representations within the identity matrix?
As other commenters have observed you have written the matrix $\lambda I$. This is useful because you will want to compute $A-\lambda I$ to find eigensystems. I.e., with your given $A$, $$ A - \lambda I = \begin{bmatrix} 1 & 2 & 1 \\ 1 & -1 & 3 \\ 4 & 5 & 10 \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} = \begin{bmatrix} 1-\lambda & 2 & 1 \\ 1 & -1-\lambda & 3 \\ 4 & 5 & 10-\lambda \end{bmatrix} $$
To find eigenvalues, we want $\det(A - \lambda I)$. \begin{align} \det(A - \lambda I) & = \det \begin{bmatrix} 1-\lambda & 2 & 1 \\ 1 & -1-\lambda & 3 \\ 4 & 5 & 10-\lambda \end{bmatrix} \\ &= (1-\lambda)(15-(-1-\lambda)(10-\lambda) ) - 2((10-\lambda)-12) + 1(5 - 4(-1-\lambda) \\ &= -12 + 22\lambda + 10 \lambda^2 - \lambda^3 \end{align} Setting this last polynomial equal to zero, $0 = -12 + 22\lambda + 10 \lambda^2 - \lambda^3$ and finding its roots, we get three horrible expressions which can be evaluated to give values near $0.455, -2.236, 11.78$. These three horrible expressions are the eigenvalues of your $A$ (and are approximated by the values given).