If a map has enough eigenvectors to make a basis, then in that basis it does nothing but stretch each basis vector by its eigenvalue, and its matrix is diagonal. Diagonalising is finding that basis, and it is possible exactly when no eigenvalue is short of eigenvectors.

A tangled knot of ropes, each pulled in its own direction, is hard to describe until you look along each rope in turn. Along each one the pull is a single number, and the whole tangle is just those numbers.

An $n\times n$ matrix $A$ is diagonalisable when $A=PDP^{-1}$ for an invertible $P$ and a diagonal $D$. Reading $AP=PD$ column by column, $A\mathbf{p}_i=d_i\mathbf{p}_i$: the columns of $P$ are eigenvectors and the diagonal of $D$ holds their eigenvalues, in the same order. So $A$ is diagonalisable exactly when $\mathbb{R}^n$ has a basis of eigenvectors of $A$, and then $D$ is the matrix of the map in that basis.

• $n$ distinct real eigenvalues make $A$ diagonalisable: their eigenvectors are independent, the theorem below, and $n$ independent vectors are a basis.
• In general $A$ is diagonalisable exactly when its characteristic polynomial has $n$ real roots counted with multiplicity and $\dim E_\mu=a_\mu$ for every eigenvalue: the eigenspaces then supply $n$ independent eigenvectors.
• The shear $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ is not diagonalisable: its one eigenvalue has $a_1=2$ and .
• The quarter turn is not diagonalisable over the real numbers: it has no real eigenvalue.
• A repeated eigenvalue does not prevent it: $I$, and every diagonal matrix, is diagonalisable.

Suppose not, and take a relation among the eigenvectors with as few non-zero weights as possible. Apply $A$, which scales each eigenvector by its eigenvalue, and subtract the relation times one of the eigenvalues: one term cancels and the rest keep non-zero weights, because the eigenvalues differ. That is a shorter relation, which cannot exist. So there was no relation at all.

Suppose the eigenvectors are dependent, and take a relation with as few non-zero weights as possible; drop the vectors with weight $0$, so every $c_i\neq0$, and $k\ge2$ because an eigenvector is not $\mathbf{0}$.

Apply $A$: each eigenvector is scaled by its own eigenvalue.

Subtract $\lambda_k$ times the first relation: the term in $\mathbf{v}_k$ cancels.

The eigenvalues are distinct and the weights non-zero, so this is a relation with fewer non-zero weights, which was ruled out.

No relation can exist, so the eigenvectors are independent.

Put a basis of eigenvectors in as the columns of $P$, and their eigenvalues, in the same order, on the diagonal of $D$. Then $AP=PD$, so $A=PDP^{-1}$.

• $A=PDP^{-1}$: eigenvectors as the columns of $P$, eigenvalues on the diagonal of $D$, in the same order.
• Diagonalisable exactly when there is a basis of eigenvectors.
• Distinct eigenvalues always do it; a repeated eigenvalue needs its eigenspace to be as large as its multiplicity.
• Shears and rotations are the typical failures.

Great work! You now know how to:

• diagonalise a matrix as $PDP^{-1}$ with matching order;
• prove that distinct eigenvalues give independent eigenvectors;
• decide diagonalisability from the eigenspaces;
• recognise shears and rotations as the typical failures.