Relation between eigenvalues and matrix's characteristic

Question

If I let $A$ be a $4×4$ real matrix with eigenvalues $−1, 1, 2, 3$ I am having hard time figuring out matrix characteristic!

1. Is matrix A invertible? (I think it is invertible since it has no $0$ eigenvalues)

2. Is A diagonalizable?(I am not sure but I think it is diagonalizable since all four eigenvalues are different. )

3. Is there a basis, for $\mathbb{R}^4$ consisting of eigenvectors of $A$? (I could not understand this question )

4. Find the determinant of $AA^{t}$ (I think it is $36$)

5. Find the trace of $(A^2 − 3I)−1$ (Is it $1/6$ right?)

Answer 1

A matrix is diagonalizable if and only if the eigen vectors of $A$ form a basis of $\mathbb{F}^{n}$.

In that suppose the matrix is of dimension $n$.

You take the matrix $P$ whose columns are those $n$ eigen vectors.

Then $P^{-1}AP=diag(\lambda_{1},\lambda_{2},...\lambda_{n})$ where $\lambda_{i}$'s are not necessarily distinct.

Now since the matrix has $4$ eigen values and eigen vectors corresponding to different eigen values are linearly independent so you have a set of $4$ linearly independent eigen vectors which is a basis as $\dim(\mathbb{R}^{4})=4$ . Write these 4 eigen vectors as the columns of $P$ to get the required transformation wrt which you get the diagonal matrix.

Also for the last part see that $Ax=\lambda x\implies A^{2}x=\lambda Ax=\lambda^{2}x$ . So $A^{2}$ has eigen values which are the squares of eigen values of $A$. And sum of eigen values is the trace of matrix. So $Tr(A^{2}-3I)=1+1+4+9-(3+3+3+3)=15-12=3$.