Tutorials › Linear Algebra › Computing the Inverse of an Eigenvector Matrix

Diagonalization · Tutorial 534 of 1000

Computing the Inverse of an Eigenvector Matrix

Compute an eigenvector matrix’s inverse and use it to translate vectors between standard coordinates and coordinates in the chosen eigenvector basis.

Intermediate 9 min read

What You'll Learn

  • Set up the augmented matrix used to compute an eigenvector matrix inverse.
  • Use row reduction to find the inverse exactly.
  • Check a computed inverse by multiplying it by the original matrix.
  • Explain how an eigenvector matrix converts basis coordinates to standard coordinates.
  • Use the inverse to recover eigenvector-basis coordinates from a vector.
  • Recognize why an eigenvector matrix’s transpose is not generally its inverse.

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.

Definition: If the columns of a square matrix \(\mathbf{P}\) form an ordered basis \(\mathcal{B}\), then \(\mathbf{P}\) is the basis matrix for \(\mathcal{B}\). For any \(\mathbf{x}\in\mathbb{R}^n\), its coordinate vector \([\mathbf{x}]_{\mathcal{B}}\) is the column of coefficients in the unique expansion \(\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n\). Thus \(\mathbf{P}[\mathbf{x}]_{\mathcal{B}}=\mathbf{x}\).

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.

Theorem: Suppose \(\mathbf{P}\in\mathbb{R}^{n\times n}\) has columns forming a basis of \(\mathbb{R}^n\). Row-reducing \(\begin{bmatrix}\mathbf{P}&\mathbf{I}_n\end{bmatrix}\) until the left block is \(\mathbf{I}_n\) gives \(\begin{bmatrix}\mathbf{I}_n&\mathbf{P}^{-1}\end{bmatrix}\).

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:

$$ \begin{bmatrix}1&1&1&0\\1&-1&0&1\end{bmatrix} \sim \begin{bmatrix}1&1&1&0\\0&-2&-1&1\end{bmatrix} \sim \begin{bmatrix}1&1&1&0\\0&1&\frac12&-\frac12\end{bmatrix} \sim \begin{bmatrix}1&0&\frac12&\frac12\\0&1&\frac12&-\frac12\end{bmatrix}. $$

Therefore \(\mathbf{P}^{-1}=\begin{bmatrix}\tfrac12&\tfrac12\\\tfrac12&-\tfrac12\end{bmatrix}\). The coordinate column is

$$ [\mathbf{x}]_{\mathcal{B}} =\mathbf{P}^{-1}\mathbf{x} =\begin{bmatrix}\frac12&\frac12\\\frac12&-\frac12\end{bmatrix} \begin{bmatrix}3\\1\end{bmatrix} =\begin{bmatrix}2\\1\end{bmatrix}. $$

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 \(R_1\):

$$ \begin{bmatrix} 1&1&0&1&0&0\\ 0&1&1&0&1&0\\ 1&0&1&0&0&1 \end{bmatrix} \sim \begin{bmatrix} 1&1&0&1&0&0\\ 0&1&1&0&1&0\\ 0&0&2&-1&1&1 \end{bmatrix} \sim \begin{bmatrix} 1&1&0&1&0&0\\ 0&1&1&0&1&0\\ 0&0&1&-\frac12&\frac12&\frac12 \end{bmatrix} $$
$$ \sim \begin{bmatrix} 1&1&0&1&0&0\\ 0&1&0&\frac12&\frac12&-\frac12\\ 0&0&1&-\frac12&\frac12&\frac12 \end{bmatrix} \sim \begin{bmatrix} 1&0&0&\frac12&-\frac12&\frac12\\ 0&1&0&\frac12&\frac12&-\frac12\\ 0&0&1&-\frac12&\frac12&\frac12 \end{bmatrix}. $$

Hence \(\mathbf{P}^{-1}=\tfrac12\begin{bmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{bmatrix}\). Multiplying by the given vector gives

$$ [\mathbf{x}]_{\mathcal{B}} =\mathbf{P}^{-1}\mathbf{x} =\begin{bmatrix} \frac12&-\frac12&\frac12\\ \frac12&\frac12&-\frac12\\ -\frac12&\frac12&\frac12 \end{bmatrix} \begin{bmatrix}2\\1\\3\end{bmatrix} =\begin{bmatrix}2\\0\\1\end{bmatrix}. $$

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 \(R_1\):

$$ \begin{bmatrix}1&2&1&0\\1&1&0&1\end{bmatrix} \sim \begin{bmatrix}1&2&1&0\\0&-1&-1&1\end{bmatrix} \sim \begin{bmatrix}1&2&1&0\\0&1&1&-1\end{bmatrix} \sim \begin{bmatrix}1&0&-1&2\\0&1&1&-1\end{bmatrix}. $$

It follows that \(\mathbf{P}^{-1}=\begin{bmatrix}-1&2\\1&-1\end{bmatrix}\), and

$$ [\mathbf{x}]_{\mathcal{B}} =\mathbf{P}^{-1}\mathbf{x} =\begin{bmatrix}-1&2\\1&-1\end{bmatrix} \begin{bmatrix}3\\2\end{bmatrix} =\begin{bmatrix}1\\1\end{bmatrix}. $$

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

Theorem: Let \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\) have the ordered basis \(\mathcal{B}\) as its columns. For every \(\mathbf{x}\in\mathbb{R}^n\), \(\mathbf{x}=\mathbf{P}[\mathbf{x}]_{\mathcal{B}}\) and \([\mathbf{x}]_{\mathcal{B}}=\mathbf{P}^{-1}\mathbf{x}\).

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.

  1. For \(\mathbf{P}=\begin{bmatrix}1&0\\2&1\end{bmatrix}\), compute \(\mathbf{P}^{-1}\).
  2. Using that same matrix, find the coordinates of \(\begin{bmatrix}3\\5\end{bmatrix}\) in the basis given by the columns of \(\mathbf{P}\).
  3. Why must each row operation be applied to both blocks of \(\begin{bmatrix}\mathbf{P}&\mathbf{I}\end{bmatrix}\)?
  4. If the columns of \(\mathbf{P}\) are reordered, what must happen to the order of the coordinates they represent?
  5. Give a multiplication check showing whether a proposed matrix \(\mathbf{Q}\) is \(\mathbf{P}^{-1}\).

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