How do I explain this in a correct way?

Question

So I was told that I have a $5\times 5$ matrix and one eigenvalue is $2$-dimensional and the second eigenvalue is $3$-dimensional and I have to explain if it's diagonalizable or not. I know that it is, since the dimensions of the 2 eigenspaces add up to 5, but I'm unsure of how I should explain in my response that it's the correct answer. It doesn't need to be a formal or anything this is an introductory linear algebra class. I want to say "Yes because the dimensions of the eigenspaces will fill $\mathbb R^5$." But I'm not sure if that makes sense. I would say that as my ansewr becasue my TA said something similar.

Answer 1

You can just say that the matrix is diagonalisable because the set of eigenvectors span $\mathbb{R}^5$.

Let $A$ denote your matrix, $\lambda_1,\lambda_2$ denote the two eigenvectors, $v_1,v_2$ ($v_3,v_4,v_5$) denote the eigenvectors corresponding to $\lambda_1$ ($\lambda_2$), $V$ be the matrix of eigenvectors and $\Lambda$ that of eigenvalues.

The above is important because it allows us to express any vector $x\in\mathbb{R}^5$ as a linear combination of the eigenvectors:

$$x=c_1v_1+c_2v_2+\dots+c_5v_5.$$

Re-arranging the above we have that $x=Vc\Leftrightarrow c=V^{-1}x$. Thus,

$$Ax=A(c_1v_1+c_2v_2+\dots+c_5v_5)=\lambda_1c_1v_1+\lambda_1c_2v_2+\dots+\lambda_2c_5v_5=V\Lambda c=V\Lambda V^{-1}x.$$

Since the above holds for all $x\in\mathbb{R}^5$ we have that $A=V\Lambda V^{-1}$.