Change of Basis and Similarity

Put basis vectors in columns

Let $B=(\mathbf b_1,\ldots,\mathbf b_n)$ be a basis and let $P$ contain those vectors as columns in standard coordinates. Then $\mathbf x=P[\mathbf x]_B$ , and $[\mathbf x]_B=P^{-1}\mathbf x$ . The direction of this conversion matters: $P$ builds the physical vector from its basis coordinates.

If a linear map has standard matrix $A$ , its matrix in basis $B$ is $A_B=P^{-1}AP$ . Follow the operations from right to left: convert to standard coordinates, apply the map, convert back. Matrices connected this way are similar and describe the same linear operator in different bases.

Visual guide

VISUAL GUIDESame vector, different coordinate labels

Standard-coordinate path

With basis b₁ = (1, 0), b₂ = (1, 1), the vector (3, 2) has basis coordinates (1, 2). One b₁ step plus two b₂ steps reaches the same geometric endpoint as three horizontal and two vertical standard steps.

Worked example: convert a vector

With $\mathbf b_1=(1,1)^T$ , $\mathbf b_2=(1,-1)^T$ , we have $P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ . For standard vector $(3,1)^T$ , solving $P\mathbf c=(3,1)^T$ gives $\mathbf c=(2,1)^T$ . Multiplying $P(2,1)^T$ verifies the conversion.

Worked example: simplify a transformation

Let $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ . Applying $A$ to the basis vectors above gives $A\mathbf b_1=3\mathbf b_1$ and $A\mathbf b_2=\mathbf b_2$ . Hence in this basis the matrix is $A_B=\operatorname{diag}(3,1)$ .

This is the idea behind diagonalization: choose directions that the transformation only scales. Such a basis does not exist for every matrix. Similarity preserves eigenvalues, determinant and trace, but it need not preserve the lengths of coordinate vectors unless the basis change is orthogonal.

For different input and output bases of a map, the formula is $Q^{-1}AP$ , where $P$ converts input coordinates and $Q$ converts output coordinates. Similarity is the special case where the same basis change is used on both sides.

Practice

If $P=\operatorname{diag}(2,3)$ , convert basis coordinates $(1,2)^T$ to standard coordinates.
With that $P$ , convert standard vector $(4,3)^T$ to basis coordinates.
Why must $P$ be invertible?

Show worked solutions

$P(1,2)^T=(2,6)^T$ .
$P^{-1}(4,3)^T=(2,1)^T$ .
Its columns must be a basis, so every vector has unique coordinates. Dependent columns would destroy uniqueness.