5.4 Eigenvectors and linear transformations

Big idea

A linear transformation between finite-dimensional spaces has no matrix until you choose bases. Choose them and every such transformation becomes a matrix, by a recipe with one moving part: apply the transformation to each basis vector of the source, then write the result in coordinates relative to the basis of the target. Those coordinate columns, side by side, are the matrix.

Different choices give different matrices for the same transformation. That is the point rather than a nuisance. Some choices produce a matrix you can read at a glance; others produce a page of arithmetic that hides what the transformation does. Choosing well is a real design decision.

Diagonalization now reads as the best possible choice. If $A = PDP^{-1}$, then $D$ is the matrix of the transformation $\mathbf{x} \mapsto A\mathbf{x}$ relative to the basis formed by the columns of $P$. Similar matrices are not merely related by an algebraic identity; they are one transformation seen from two coordinate systems, and the change of coordinates is $P$.

Decoder

The $\mathcal{B}$-matrix of $T$ has as its $j$th column the $\mathcal{B}$-coordinate vector of $T(\mathbf{b}_j)$.

Slowly: take the $j$th vector of your basis, push it through $T$, then express the output as a combination of the same basis vectors. The weights in that combination, stacked into a column, form column $j$. Do this for each basis vector in turn and you have the whole matrix. Every “matrix of a transformation” formula in the book is that instruction with different names attached.

Definitions and results

Coordinates. Let $\mathcal{B} = \{\mathbf{b}_1, \dots, \mathbf{b}_n\}$ be a basis of $V$. Every $\mathbf{x}$ in $V$ is $c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n$ for exactly one list of weights, and the column $[\mathbf{x}]_{\mathcal{B}} = (c_1, \dots, c_n)$ is the coordinate vector of $\mathbf{x}$. Coordinates respect sums and scalar multiples, which is what lets a matrix represent $T$ at all.

Change of coordinates in $\mathbb{R}^n$. Let $P_{\mathcal{B}}$ be the matrix whose columns are the basis vectors themselves. Then $\mathbf{x} = P_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}$, so $P_{\mathcal{B}}$ converts $\mathcal{B}$-coordinates into standard coordinates and $P_{\mathcal{B}}^{-1}$ goes the other way. Read the arrow direction carefully; getting it backwards is the most common error in this section.

The matrix of $T$ relative to two bases. For $T: V \to W$ with basis $\mathcal{B}$ for $V$ and $\mathcal{C}$ for $W$, the matrix $M$ has $j$th column $[T(\mathbf{b}_j)]_{\mathcal{C}}$, and then $[T(\mathbf{x})]_{\mathcal{C}} = M[\mathbf{x}]_{\mathcal{B}}$ for every $\mathbf{x}$. When $V = W$ and one basis is used on both sides, $M$ is written $[T]_{\mathcal{B}}$ and called the $\mathcal{B}$-matrix of $T$.

The standard matrix is a special case. For $T: \mathbb{R}^n \to \mathbb{R}^m$ with the standard bases, the recipe gives the familiar matrix whose columns are $T(\mathbf{e}_1), \dots, T(\mathbf{e}_n)$.

Similarity as a change of coordinates. Let $A$ be $n \times n$, let $\mathcal{B}$ be a basis of $\mathbb{R}^n$, and let $T(\mathbf{x}) = A\mathbf{x}$. Then

$$ [T]_{\mathcal{B}} = P_{\mathcal{B}}^{-1} A P_{\mathcal{B}} $$

Read the right side from right to left: convert $\mathcal{B}$-coordinates to standard, apply $A$, convert back. Two similar matrices are therefore the same transformation written in two bases, and any property that does not depend on the coordinate system, such as the eigenvalues, the determinant, the trace, and the rank, is shared by both.

Diagonal matrix representation. $A$ is diagonalizable exactly when there is a basis of $\mathbb{R}^n$ in which the transformation $\mathbf{x} \mapsto A\mathbf{x}$ has a diagonal matrix. That basis is a basis of eigenvectors, and the diagonal entries are the eigenvalues. If $A = PDP^{-1}$, the basis is the set of columns of $P$ and $D = [T]_{\mathcal{B}}$.

Eigenvectors of a transformation. For a transformation on any vector space, not just $\mathbb{R}^n$, a nonzero $\mathbf{v}$ with $T(\mathbf{v}) = \lambda\mathbf{v}$ is an eigenvector. Polynomial spaces and spaces of matrices have eigenvectors in exactly this sense, and their $\mathcal{B}$-matrices are handled with the same recipe.

Dimension is forced. If $V$ has dimension $n$ and $W$ has dimension $m$, then the matrix of $T: V \to W$ is $m \times n$ for any choice of bases. Choice affects the entries, never the shape.

Worked examples

The matrix in an eigenvector basis

Let $T(\mathbf{x}) = A\mathbf{x}$ with

$$ A = \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} $$

and take $\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\}$ with $\mathbf{b}_1 = (1,1)$ and $\mathbf{b}_2 = (2,1)$, the eigenvectors for $\lambda = 2$ and $\lambda = 3$.

Apply the recipe. $T(\mathbf{b}_1) = 2\mathbf{b}_1$, so its $\mathcal{B}$-coordinates are $(2,0)$. $T(\mathbf{b}_2) = 3\mathbf{b}_2$, so its $\mathcal{B}$-coordinates are $(0,3)$. Therefore

$$ [T]_{\mathcal{B}} = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} $$

Now run a vector through both descriptions. Take $\mathbf{x} = (8,5)$. Its $\mathcal{B}$-coordinates solve $c_1(1,1) + c_2(2,1) = (8,5)$, that is $c_1 + 2c_2 = 8$ and $c_1 + c_2 = 5$, giving $c_2 = 3$ and $c_1 = 2$. Multiplying by the diagonal matrix gives $[T(\mathbf{x})]_{\mathcal{B}} = (4, 9)$. Converting back, $4(1,1) + 9(2,1) = (22, 13)$.

Check against the direct route: $A(8,5) = (32 - 10, \; 8 + 5) = (22, 13)$. The two agree, and the diagonal route needed only two multiplications.

The same transformation in a different basis

Keep $A$ and take $\mathcal{B} = \{(1,0), (1,1)\}$, which contains no eigenvector. Then

$$ P_{\mathcal{B}} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \qquad P_{\mathcal{B}}^{-1} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} $$

Compute $A P_{\mathcal{B}}$ first, then apply $P_{\mathcal{B}}^{-1}$:

$$ A P_{\mathcal{B}} = \begin{bmatrix} 4 & 2 \\ 1 & 2 \end{bmatrix}, \qquad [T]_{\mathcal{B}} = P_{\mathcal{B}}^{-1} A P_{\mathcal{B}} = \begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix} $$

Check with the column recipe instead of the formula. $T(1,0) = (4,1)$, and writing $(4,1) = c_1(1,0) + c_2(1,1)$ forces $c_2 = 1$ and $c_1 = 3$, so the first column is $(3,1)$. $T(1,1) = (2,2)$, which is $0(1,0) + 2(1,1)$, so the second column is $(0,2)$. Both routes give the same matrix.

The result is triangular, not diagonal, but its diagonal entries are $3$ and $2$: the eigenvalues survive the change of basis even though this basis was not built from eigenvectors.

A transformation on polynomials

Let $V$ be the space of polynomials of degree at most $2$, and define $T(p)(t) = p(t) + t\,p'(t)$. Use the basis $\mathcal{B} = \{1, t, t^2\}$.

Apply $T$ to each basis vector. For $p = 1$, the derivative is $0$, so $T(1) = 1$. For $p = t$, $T(t) = t + t(1) = 2t$. For $p = t^2$, $T(t^2) = t^2 + t(2t) = 3t^2$. Reading coordinates relative to $\mathcal{B}$:

$$ [T]_{\mathcal{B}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} $$

Diagonal, so the basis vectors are eigenvectors of $T$ with eigenvalues $1$, $2$, $3$. Check one against the definition on a mixed polynomial: for $p(t) = 4 + 5t - 2t^2$, the recipe gives $T(p) = 4 + 10t - 6t^2$, while the matrix gives coordinates $(1 \cdot 4, \, 2 \cdot 5, \, 3 \cdot (-2)) = (4, 10, -6)$. Same polynomial.

Practice

Build the matrix of a transformation from the images of the basis vectors.

Convert between standard coordinates and coordinates relative to a chosen basis, in both directions.

Then connect the two: find the basis that makes a matrix diagonal, and write the factorization.

Videos

Watch how a vector gets two different coordinate lists depending on who is describing it, and how translating a transformation into someone else’s coordinates produces the sandwich $P^{-1}AP$. That sandwich is the formula in this section.

Quiz

Six items on coordinate vectors, matrices of transformations, and diagonal representations.