How to prove each column vector $t^i$ of $\Bbb{R}^n$ satisfies $At^i = \lambda(t^i)$?

Question

If we have a matrix $A \in \mathbb{R}^{n \times n}$ and $A=T\Lambda T^{-1}$ for $T\in \mathbb{R}^{n \times n}$, how to prove that each column vector $t^i \in \mathbb{R}^n$ that $At^i = \lambda(t^i)$?

I'm having trouble understanding how to prove that each element $t^i$ of $T$ is still an eigenvalue/vector pair of $A$. Thus saying that $(t^i, \lambda_i)$ is an eigenvalue and eigenvector pair since $A \in \mathbb{R}^{n \times n}$ but $t$ is a column vector of $T$ and is only $\mathbb{R}^n$.

Any help in proving this would be much appreciated.

Answer 1

$$A=T\Lambda T^{-1}$$

Multiply $T$ on both sides from the right.

$$AT=T\Lambda$$

Look at the $i$ column or if it helps, multiply by $e_i$ where $e_i$ is the $i$-th standard unit vector, there you can observe that

$$At^i = \lambda_i t^i$$

Answer 2

So $A$ and $T$ are both $n \times n$ matrices. Note that any eigenvalue/eigenvector pair $\vec{x}, \lambda$ must satisfy $$ A \vec{x} = \lambda \vec{x} \quad (*) $$ and so $\vec{x} \in \mathbb{R}^n$. If you take $k$ such pairs $\left(\vec{t}_i, \lambda_i\right)_{i=1}^k$, each one satisfying an equation like $(*)$, you can unite all such equations into a matrix equation $$ AT = T \Lambda, $$ where $T = (t_1| \ldots| t_k)$ will be $n \times k$ and $\Lambda$ will be a diagonal matrix with diagonal values $\lambda_1, \ldots, \lambda_k$.

When $k=n$ you can show $T$ is invertible, so you get $$ A = T \Lambda T^{-1} $$ as in your problem.