Why Compute the Inverse of an Eigenvector Matrix?
When a matrix is diagonalizable, its chosen eigenvectors can be placed in the columns of a matrix \(\mathbf{P}\). The order of those columns is important: it fixes the order of the basis vectors and must match the order of the corresponding eigenvalues. Computing \(\mathbf{P}^{-1}\) gives a practical way to change coordinates. The matrix \(\mathbf{P}\) turns coordinates in the eigenvector basis into standard coordinates; its inverse carries a vector back to its coordinates in that basis.
Throughout, vectors and matrices have real entries. An ordered basis \(\mathcal{B}=(\mathbf{v}_1,\dots,\mathbf{v}_n)\) of \(\mathbb{R}^n\) is a list that is linearly independent and spans \(\mathbb{R}^n\). If \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\), then multiplying \(\mathbf{P}\) by a column of coefficients forms the corresponding linear combination of the basis vectors. The key task is to find the matrix that reverses this operation.
In a diagonalization, the columns of \(\mathbf{P}\) are eigenvectors, and each diagonal entry of the diagonal matrix is the eigenvalue belonging to the eigenvector in the corresponding column. The earlier result that a matrix is diagonalizable exactly when it has an eigenbasis ensures that such a square basis matrix is invertible. This tutorial focuses on computing its inverse and interpreting the coordinate change, rather than testing or verifying a diagonalization.
Row Reduction for the Inverse
To compute \(\mathbf{P}^{-1}\), augment \(\mathbf{P}\) with the identity matrix and row-reduce the left side to the identity. Apply every row operation to the entire augmented matrix, including the right-hand side. The block that ends up on the right is \(\mathbf{P}^{-1}\). This method works when \(\mathbf{P}\) is invertible; here, that condition is met because its columns are a basis.
Proof. For each column \(\mathbf{e}_j\) of \(\mathbf{I}_n\), the basis property gives a unique coefficient column \(\mathbf{c}_j\) such that \(\mathbf{P}\mathbf{c}_j=\mathbf{e}_j\). Collect these columns into \(\mathbf{C}=\begin{bmatrix}\mathbf{c}_1&\cdots&\mathbf{c}_n\end{bmatrix}\). Then \(\mathbf{P}\mathbf{C}=\mathbf{I}_n\), column by column. Row operations on an augmented matrix perform the same operations on each right-hand-side column and preserve the solution sets of the corresponding systems. Reducing the left block to \(\mathbf{I}_n\) therefore turns each system \(\mathbf{P}\mathbf{c}_j=\mathbf{e}_j\) into \(\mathbf{I}_n\mathbf{c}_j=\mathbf{c}_j\). The right block after reduction has columns \(\mathbf{c}_1,\dots,\mathbf{c}_n\), so it is \(\mathbf{C}\). Since \(\mathbf{P}\mathbf{C}=\mathbf{I}_n\), \(\mathbf{C}\) is the inverse of \(\mathbf{P}\), as required. \(\square\)
There are two useful checks. First, multiply the proposed inverse on the left by \(\mathbf{P}\) and confirm the result is \(\mathbf{I}_n\). Second, for a specific vector \(\mathbf{x}\), calculate \(\mathbf{c}=\mathbf{P}^{-1}\mathbf{x}\), then check that \(\mathbf{P}\mathbf{c}=\mathbf{x}\). The first checks the matrix calculation; the second checks the coordinate interpretation in a particular case.
Worked Inverses and Coordinate Changes
Worked Example: A Two-Vector Eigenbasis
Suppose the ordered eigenbasis is \(\mathcal{B}=(\mathbf{v}_1,\mathbf{v}_2)\), with \(\mathbf{v}_1=\begin{bmatrix}1\\1\end{bmatrix}\) and \(\mathbf{v}_2=\begin{bmatrix}1\\-1\end{bmatrix}\). Compute the basis matrix inverse and find the \(\mathcal{B}\)-coordinates of \(\mathbf{x}=\begin{bmatrix}3\\1\end{bmatrix}\).
Place the basis vectors in the columns, then augment with the identity. Subtract the first row from the second, scale the second row by \(-\tfrac12\), and subtract the new second row from the first:
Therefore \(\mathbf{P}^{-1}=\begin{bmatrix}\tfrac12&\tfrac12\\\tfrac12&-\tfrac12\end{bmatrix}\). The coordinate column is
Check: The claimed coordinates say \(\mathbf{x}=2\mathbf{v}_1+\mathbf{v}_2\). Indeed, \(2\begin{bmatrix}1\\1\end{bmatrix}+\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}\). Also, direct multiplication gives \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}_2\).
Worked Example: A Three-Vector Eigenbasis
Suppose three chosen eigenvectors are the columns of \(\mathbf{P}=\begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}\). Compute \(\mathbf{P}^{-1}\) and find the coordinates of \(\mathbf{x}=\begin{bmatrix}2\\1\\3\end{bmatrix}\) in this ordered basis.
Begin with the augmented matrix. Subtract \(R_1\) from \(R_3\), add \(R_2\) to the new \(R_3\), and scale \(R_3\) by \(\tfrac12\). Then subtract \(R_3\) from \(R_2\) and subtract \(R_2\) from \(R_1\):
Hence \(\mathbf{P}^{-1}=\tfrac12\begin{bmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{bmatrix}\). Multiplying by the given vector gives
Check: The coordinates give \(2\mathbf{v}_1+0\mathbf{v}_2+\mathbf{v}_3\), where \(\mathbf{v}_1=\begin{bmatrix}1\\0\\1\end{bmatrix}\) and \(\mathbf{v}_3=\begin{bmatrix}0\\1\\1\end{bmatrix}\). Thus \(2\mathbf{v}_1+\mathbf{v}_3=\begin{bmatrix}2\\1\\3\end{bmatrix}=\mathbf{x}\). Multiplying the displayed inverse by \(\mathbf{P}\) also gives \(\mathbf{I}_3\).
Worked Example: A Non-Orthogonal Eigenvector Basis
Let the ordered basis vectors be the columns of \(\mathbf{P}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\). Compute the inverse and use it to find the coordinates of \(\mathbf{x}=\begin{bmatrix}3\\2\end{bmatrix}\).
Subtract \(R_1\) from \(R_2\), multiply \(R_2\) by \(-1\), and subtract twice the new \(R_2\) from \(R_1\):
It follows that \(\mathbf{P}^{-1}=\begin{bmatrix}-1&2\\1&-1\end{bmatrix}\), and
Check: With \(\mathbf{v}_1=\begin{bmatrix}1\\1\end{bmatrix}\) and \(\mathbf{v}_2=\begin{bmatrix}2\\1\end{bmatrix}\), the coordinates \(\begin{bmatrix}1\\1\end{bmatrix}\) give \(\mathbf{v}_1+\mathbf{v}_2=\begin{bmatrix}3\\2\end{bmatrix}\). The inverse check also holds: \(\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}-1&2\\1&-1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\).
How the Two Coordinate Directions Work
Proof. By the definition of the coordinate vector, \(\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n\), where \([\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}c_1&\cdots&c_n\end{bmatrix}^{\top}\). By the definition of matrix multiplication, \(\mathbf{P}[\mathbf{x}]_{\mathcal{B}}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n=\mathbf{x}\). Since \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}_n\), multiplying the equality \(\mathbf{x}=\mathbf{P}[\mathbf{x}]_{\mathcal{B}}\) by \(\mathbf{P}^{-1}\) gives \(\mathbf{P}^{-1}\mathbf{x}=[\mathbf{x}]_{\mathcal{B}}\). \(\square\)
The direction matters. A coordinate column such as \(\begin{bmatrix}2\\0\\1\end{bmatrix}\) lists coefficients relative to the chosen basis; multiplying by \(\mathbf{P}\) builds the corresponding vector in standard coordinates. In contrast, \(\mathbf{P}^{-1}\mathbf{x}\) starts with a vector in standard coordinates and returns its coefficients relative to the ordered basis. These are inverse operations, not interchangeable ones.
For a diagonalization, the column order must be kept consistent throughout. If the first column of \(\mathbf{P}\) is an eigenvector for \(\lambda_1\) and the second is an eigenvector for \(\lambda_2\), then the first and second coordinates returned by \(\mathbf{P}^{-1}\) are coefficients on those same eigenvectors, in that order. Reordering columns changes the coordinate system and requires recomputing or correspondingly reordering the inverse.
Common Mistake: Using the Transpose
A common error is to assume that \(\mathbf{P}^{-1}=\mathbf{P}^{\top}\) for every eigenvector matrix. The transpose is the inverse when the columns are orthonormal, but eigenvectors in a basis need not be orthonormal. For the last example, a wrong calculation would use \(\mathbf{P}^{\top}=\begin{bmatrix}1&1\\2&1\end{bmatrix}\). It gives \(\mathbf{P}^{\top}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}5\\8\end{bmatrix}\), which is not the coordinate vector: multiplying back gives \(\mathbf{P}\begin{bmatrix}5\\8\end{bmatrix}=\begin{bmatrix}21\\13\end{bmatrix}\ne\begin{bmatrix}3\\2\end{bmatrix}\). The correction is to row-reduce \(\begin{bmatrix}\mathbf{P}&\mathbf{I}\end{bmatrix}\), giving \(\mathbf{P}^{-1}=\begin{bmatrix}-1&2\\1&-1\end{bmatrix}\) and coordinates \(\begin{bmatrix}1\\1\end{bmatrix}\), which pass the reconstruction check.
Before using an inverse in a diagonalization calculation, make these checks:
- Confirm that the columns of \(\mathbf{P}\) are in the intended eigenvector order.
- Apply every row operation to both blocks of the augmented matrix.
- Multiply \(\mathbf{P}\mathbf{P}^{-1}\) to check the inverse.
- For a coordinate calculation, reconstruct the vector by multiplying the coordinates by \(\mathbf{P}\).
These checks distinguish a correct inverse from a plausible-looking matrix and confirm that the entries have the intended coordinate meaning.
Check Your Understanding
Use row reduction and multiplication to check your results.
- For \(\mathbf{P}=\begin{bmatrix}1&0\\2&1\end{bmatrix}\), compute \(\mathbf{P}^{-1}\).
- Using that same matrix, find the coordinates of \(\begin{bmatrix}3\\5\end{bmatrix}\) in the basis given by the columns of \(\mathbf{P}\).
- Why must each row operation be applied to both blocks of \(\begin{bmatrix}\mathbf{P}&\mathbf{I}\end{bmatrix}\)?
- If the columns of \(\mathbf{P}\) are reordered, what must happen to the order of the coordinates they represent?
- Give a multiplication check showing whether a proposed matrix \(\mathbf{Q}\) is \(\mathbf{P}^{-1}\).