Linear Algebra: Proving Matrices 
Let $\boldsymbol{A}$ and $\boldsymbol{P} = \begin{pmatrix}\boldsymbol{u} & \boldsymbol{v} & \boldsymbol{w}\end{pmatrix}$ be $3\times 3$ matrices where $\boldsymbol{u},$ $\boldsymbol{v}$ and $\boldsymbol{w}$ are columns of $\boldsymbol{P}$ such that
  $$\boldsymbol{Au}=a\boldsymbol{u},\quad \boldsymbol{Av}=b\boldsymbol{v},\quad\text{and} \quad\boldsymbol{Aw}=c\boldsymbol{w}$$
  for some real numbers $a,$ $b$ and $c.$
Show that if $\boldsymbol{P}$ is invertible then
  $\boldsymbol{A} = \boldsymbol{P}\begin{pmatrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{pmatrix}\boldsymbol{P}^{-1}.$

Can anyone help me with this problem? I am having a hard time understanding and interpreting this question as it's my first time learning about linear algebra. Thank you so much!
 A: Let $D = \begin{pmatrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{pmatrix}$. The statement $A = PDP^{-1}$ is equivalent to the statement $AP = PD$.
We now show that these two matrices are equal, by showing that their columns are equal, i.e. that $APe_1 = PDe_1,~ APe_2 = PDe_2,~ APe_3 = PDe_3$, where $e_1 = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, e_2 = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}, e_3 = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$ are the standard unit vectors.
So let's check that:
$$APe_1 = Au = au = aPe_1 = Pae_1 = PDe_1,$$
$$APe_2 = Av = bv = bPe_2 = Pbe_2 = PDe_2,$$
$$APe_3 = Aw = cw = cPe_3 = Pce_3 = PDe_3.$$
We're done. Note that $AP = PD$ is also true if $P$ is not invertible, but in that case it doesn't make sense to write $A = PDP^{-1}$.
A: Since it given that $A\mathbf{u}=a\mathbf{u}$, $A\mathbf{v}=b\mathbf{v}$ and $A\mathbf{w}=c\mathbf{w}$ for some $a,b,c\in\mathbb{R}$, we can infer that $a,b,c$  are eigen values and $\mathbf{u},\mathbf{v},\mathbf{w}$ are eigen vectors of the matrix $A$. 
Now suppose that the given matrix $P$ is invertible. Since $P$ is invertible, the column vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ are linearly independent, and thus $\mathbf{u},\mathbf{v},\mathbf{w}$ are linearly independent eigen vectors of the matrix $A$. Since $\mathbf{u},\mathbf{v},\mathbf{w}$ are linearly independent eigen vectors, the matrix $A$ is diagonalizable, and we can write 
$A=P\Lambda P^{-1}$   where $\Lambda=\begin{bmatrix} a & 0 & 0\\ 0 & b & 0\\0 & 0 & c\end{bmatrix}$
QED
