From Eigenvectors to a Factorization
Suppose a matrix has a collection of linearly independent eigenvectors. Those vectors can be assembled into a matrix, while their corresponding eigenvalues are placed on a diagonal. The resulting factorization records how the matrix acts on those vectors all at once. The order matters: each diagonal entry must belong to the eigenvector in the same column of the eigenvector matrix.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). An eigenvector \(\mathbf{v}\) of \(\mathbf{A}\) is a nonzero vector satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some real scalar \(\lambda\), its eigenvalue. A diagonalization writes \(\mathbf{A}\) as \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{P}\) is invertible and \(\mathbf{D}\) is diagonal. The construction below explains what the factors contain and why their product recovers \(\mathbf{A}\).
Why does independence matter? The columns of \(\mathbf{P}\) must give unique coordinates for every vector in \(\mathbb{R}^n\), so that \(\mathbf{P}^{-1}\) exists. For a square matrix, \(n\) independent columns imply a pivot in every column when row reduced; with \(n\) rows, this makes the matrix invertible. A list of eigenvectors that is dependent may still consist entirely of valid eigenvectors, but it cannot be used as the columns of an invertible \(\mathbf{P}\).
Why the Construction Works
Proof. The \(i\)th column of \(\mathbf{A}\mathbf{P}\) is \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\). The \(i\)th column of \(\mathbf{P}\mathbf{D}\) is also \(\lambda_i\mathbf{v}_i\), because right multiplication by the diagonal matrix multiplies column \(i\) of \(\mathbf{P}\) by \(\lambda_i\). Thus the two matrices have the same columns, so \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Independence of the \(n\) columns makes \(\mathbf{P}\) invertible. Multiplying the equation on the right by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}\mathbf{P}\mathbf{P}^{-1}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Since \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}_n\), this is \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). \(\square\)
The proof gives a useful check before calculating an inverse: confirm that each proposed column really is an eigenvector for the eigenvalue in the matching position. The matrix equation \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) tests all the eigenvector equations at once. Once it holds and \(\mathbf{P}\) is invertible, the factorization follows.
Worked Example: A Two-Column Construction
Let \(\mathbf{A}=\begin{bmatrix}4&1\\0&2\end{bmatrix}\), with eigenvectors \(\mathbf{v}_1=\begin{bmatrix}1\\0\end{bmatrix}\) for \(\lambda_1=4\) and \(\mathbf{v}_2=\begin{bmatrix}-1\\2\end{bmatrix}\) for \(\lambda_2=2\). Construct and verify \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
First check the eigenvector equations. For the first vector, \(\mathbf{A}\mathbf{v}_1=\begin{bmatrix}4\\0\end{bmatrix}=4\mathbf{v}_1\). For the second, \(\mathbf{A}\mathbf{v}_2=\begin{bmatrix}-4+2\\4\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix}=2\mathbf{v}_2\). Put the vectors in the given order and place their eigenvalues in that same order:
The columns of \(\mathbf{P}\) are independent: if \(c_1\begin{bmatrix}1\\0\end{bmatrix}+c_2\begin{bmatrix}-1\\2\end{bmatrix}=\mathbf{0}\), the second coordinate gives \(2c_2=0\), so \(c_2=0\), and then the first gives \(c_1=0\). To find the inverse, row-reduce the augmented matrix:
Thus \(\mathbf{P}^{-1}=\begin{bmatrix}1&\frac12\\0&\frac12\end{bmatrix}\). Now multiply in stages:
Check: the product agrees entry by entry with the original matrix. Also, multiplying \(\mathbf{A}\mathbf{P}\) gives \(\begin{bmatrix}4&-2\\0&4\end{bmatrix}\), which equals \(\mathbf{P}\mathbf{D}\), as the theorem predicts.
The Coordinate Meaning of the Factors
The factorization can be read as a sequence of three operations on a vector. The matrix \(\mathbf{P}^{-1}\) converts standard coordinates into coordinates relative to the ordered eigenvector list. The diagonal matrix \(\mathbf{D}\) scales each of those coordinates by the corresponding eigenvalue. Finally, \(\mathbf{P}\) converts the result back to standard coordinates. This interpretation helps explain why the eigenvector order must be consistent throughout the construction.
Proof. Since \(\mathbf{x}=\mathbf{P}\mathbf{c}\), multiply by \(\mathbf{A}\) to obtain \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{P}\mathbf{c}\). The eigenvector equations give \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), so \(\mathbf{A}\mathbf{x}=\mathbf{P}\mathbf{D}\mathbf{c}\). Because \(\mathbf{P}\) is invertible, the coordinates of \(\mathbf{A}\mathbf{x}\) in this ordered list are exactly \(\mathbf{D}\mathbf{c}\). \(\square\)
Worked Example: Building the Factors for a Three-by-Three Matrix
Consider the matrix and ordered eigenvectors \(\mathbf{A}=\begin{bmatrix}2&1&-1\\-\frac12&\frac72&\frac12\\-\frac32&\frac32&\frac52\end{bmatrix}\), \(\mathbf{v}_1=\begin{bmatrix}1\\0\\1\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}1\\1\\0\end{bmatrix}\), and \(\mathbf{v}_3=\begin{bmatrix}0\\1\\1\end{bmatrix}\), with corresponding eigenvalues \(1,3,4\). Construct the factors and check the result.
Place the vectors in the columns of \(\mathbf{P}\). The matching eigenvalues go in the diagonal positions of \(\mathbf{D}\):
The columns are independent. Indeed, if \(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\), the coordinate equations are \(c_1+c_2=0\), \(c_2+c_3=0\), and \(c_1+c_3=0\). The first two give \(c_2=-c_1\) and \(c_3=c_1\); the third then gives \(2c_1=0\), so all three coefficients are zero. To compute the inverse, multiply the following matrices:
Next, scale each column of \(\mathbf{P}\) by its matching eigenvalue to form \(\mathbf{P}\mathbf{D}\):
Multiplying this by the inverse gives
Check: multiplying \(\mathbf{A}\mathbf{P}\) gives \(\begin{bmatrix}1&3&0\\0&3&4\\1&0&4\end{bmatrix}\), exactly the displayed \(\mathbf{P}\mathbf{D}\). Its columns are \(1\mathbf{v}_1\), \(3\mathbf{v}_2\), and \(4\mathbf{v}_3\), so each eigenvector equation is verified as well.
Order, Scaling, and a Common Error
The columns of \(\mathbf{P}\) form an ordered list, not an unordered set. If two eigenvectors exchange positions, their eigenvalues must exchange positions in \(\mathbf{D}\) as well. Otherwise, the diagonal matrix scales a vector by the wrong eigenvalue. Eigenvectors may also be multiplied by nonzero scalars: this changes \(\mathbf{P}\), but it does not change the eigenvalue that belongs in the corresponding diagonal position.
Worked Example: Reordering the Same Eigenvectors
Use the two eigenvectors from the first worked example, but reverse their order. The new columns are \(\mathbf{v}_2,\mathbf{v}_1\), so set \(\mathbf{P}'=\begin{bmatrix}-1&1\\2&0\end{bmatrix}\) and match them with \(\mathbf{D}'=\begin{bmatrix}2&0\\0&4\end{bmatrix}\).
The inverse is \((\mathbf{P}')^{-1}=\begin{bmatrix}0&\frac12\\1&\frac12\end{bmatrix}\), as direct multiplication confirms: \(\begin{bmatrix}-1&1\\2&0\end{bmatrix}\begin{bmatrix}0&\frac12\\1&\frac12\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Now compute:
Check: the reversed ordering still yields the original matrix because the eigenvalues were reversed with the eigenvectors. If we instead kept \(\mathbf{D}=\operatorname{diag}(4,2)\), then \(\mathbf{P}'\mathbf{D}\) would be \(\begin{bmatrix}-4&2\\8&0\end{bmatrix}\), and multiplying by \((\mathbf{P}')^{-1}\) would give \(\begin{bmatrix}2&-1\\0&4\end{bmatrix}\), not \(\mathbf{A}\).
Common pitfall: mismatched pairs. The incorrect calculation above did not fail because either vector stopped being an eigenvector. It failed because the first column of \(\mathbf{P}'\) is an eigenvector for \(2\), while the first diagonal entry was incorrectly set to \(4\). The correction is to keep each pair together: column \(i\) of \(\mathbf{P}\) must satisfy \(\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\), and entry \(i\) of \(\mathbf{D}\) must be \(\lambda_i\).
A second warning is that merely collecting \(n\) eigenvectors does not guarantee that they can be used in this formula: they must be linearly independent. For example, for the identity matrix \(\mathbf{I}_2\), both \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}2\\0\end{bmatrix}\) are eigenvectors, but using them as columns gives a dependent list and a matrix with no inverse. Choose an independent list before computing \(\mathbf{P}^{-1}\).
Putting the Procedure into Practice
When a suitable ordered list of eigenvectors is already known, constructing and checking the factorization is systematic:
For each proposed eigenvector, record the eigenvalue that satisfies its defining equation.
Put the eigenvectors as columns of \(\mathbf{P}\), and put the matching eigenvalues on the diagonal of \(\mathbf{D}\) in the same order.
Verify that the columns of \(\mathbf{P}\) are independent. Only then does \(\mathbf{P}^{-1}\) exist.
Check \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) column by column or by matrix multiplication.
Compute \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) and compare every entry with \(\mathbf{A}\).
The intermediate check \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) is often quicker than a full product and can locate an ordering error immediately. The final product is still valuable: it verifies that the inverse and matrix multiplication have also been computed correctly. The factorization is not a new matrix unrelated to \(\mathbf{A}\); it expresses the same linear action in coordinates adapted to the chosen eigenvectors.
Check Your Understanding
For each question, keep the columns of \(\mathbf{P}\) and the diagonal entries of \(\mathbf{D}\) in matching order.
- If the columns of \(\mathbf{P}\) are eigenvectors \(\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\) for eigenvalues \(5,1,5\), write the corresponding diagonal matrix \(\mathbf{D}\).
- Explain why \(n\) eigenvectors in \(\mathbb{R}^n\) must be linearly independent for the expression \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) to be defined with those vectors as columns.
- Suppose \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) and \(\mathbf{P}\) is invertible. Show by right multiplication that \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
- For the first worked example, reverse the columns of \(\mathbf{P}\) and state the diagonal entries that must be reversed with them.
- Give one reason why checking \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) is useful before computing the full factorization.
