How to Diagonalize a Matrix: A Quick Linear Algebra Guide

Find the eigenvalues, eigenvectors, and the diagonal matrix

Written by Kyle Smith

Last Updated: February 16, 2023 References

Find the Eigenvalues

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Find the Eigenvectors

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Diagonalize the Matrix

Finding a diagonal matrix can be a lengthy process, but it’s easy if you know the steps! You’ll need to calculate the eigenvalues, get the eigenvectors for those values, and use the diagonalization equation. Diagonal matrices are great for many different operations, such as computing the powers of the matrix. This wikiHow guide shows you how to diagonalize a matrix.

Things You Should Know

Find the eigenvalues of your given matrix.
Use the eigenvalues to get the eigenvectors.
Apply the diagonalization equation using the eigenvectors to find the diagonal matrix.
Note that not all matrices can be diagonalized.

Part 1

Part 1 of 3:

Find the Eigenvalues

1
Recall the equation for finding eigenvalues. Eigenvalues are the scalar value associated with an eigenvector, represented by the symbol lambda (λ). To find eigenvalues, use the following equation:^{[1]
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- $det(lambda*I-A)=0$
- In other words, the determinant of lambda times the identity matrix minus the given transformation matrix.
2
Set up the determinant equation. Here is an example for a 2 by 2 matrix:
- $det(lambda{\begin{bmatrix}1&0\\0&1\end{bmatrix}}-{\begin{bmatrix}2&7\\8&3\end{bmatrix}})=0$
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3
Simplify the determinant equation. The equation simplifies to:
- $det({\begin{bmatrix}lambda-2&-7\\-8&lambda-3\end{bmatrix}})=0$
4
Solve for the determinant. To do so in a 2 by 2 matrix, multiply the top-left value and the bottom-right value, then subtract the product of the top-right value and bottom-left value. We also have a guide for finding the determinant for 3x3 matrices. Continuing our example:
- $(lambda-2)(lambda-3)-(-8)(-7)=0$
- $lambda^{2}-5lambda-50=0$
- $lambda=-5$ and $lambda=10$
- These two values are the eigenvalues.

Part 2

Part 2 of 3:

Find the Eigenvectors

1
Use the equation to find eigenvectors. Recall that the equation to find eigenvectors for a given lambda:^{[2]
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- $AV=lambda*V$ $AV=lambda*V$
  - In other words, a given transformation matrix (A) times the eigenvector (V) equals the eigenvalue (lambda) times the eigenvector (V).
- This can be rewritten as:
  - $lambda*V-AV=0$
  - $(lambda*I-A)V=0$
  - Where I is the identity matrix.
2
Plug in an eigenvalue to the equation. For example, using the eigenvalue lambda = -5 from our problem:
- $(-5{\begin{bmatrix}1&0\\0&1\end{bmatrix}}-{\begin{bmatrix}2&7\\8&3\end{bmatrix}})V=0$
- $({\begin{bmatrix}-5&0\\0&-5\end{bmatrix}}-{\begin{bmatrix}2&7\\8&3\end{bmatrix}})V=0$
- ${\begin{bmatrix}-7&-7\\-8&-8\end{bmatrix}}V=0$
3
Find the reduced row echelon form. Use elementary row operations to get the reduced row echelon form of the new matrix. Continuing our example:
- Our starting matrix:
  - ${\begin{bmatrix}-7&-7\\-8&-8\end{bmatrix}}V=0$
- The matrix in reduced row echelon form:
  - ${\begin{bmatrix}1&1\\0&0\end{bmatrix}}V=0$
4
Find the eigenvector. To do so:
- Our starting matrix equation:
  - ${\begin{bmatrix}1&1\\0&0\end{bmatrix}}*{\begin{bmatrix}V_{1}\\V_{2}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$
- Writing the matrix as a linear equation:
  - $V_{1}+V_{2}=0$
  - $V_{1}=-V_{2}$
- Write the components as an eigenvector:
  - ${\begin{bmatrix}-1\\1\end{bmatrix}}$
5
Repeat the above eigenvector process for any other eigenvalues. You’ll need the eigenvectors for each of the eigenvalues to diagonalize the matrix. For example, repeating the process for lambda = 10 yields the eigenvector:
- ${\begin{bmatrix}7\\8\end{bmatrix}}$

Part 3

Part 3 of 3:

Diagonalize the Matrix

1
Note the equation for diagonalizing a matrix. The equation is:^{[3]
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- P^-1 * A * P = D
- Where P is the matrix of eigenvectors, A is the given matrix, and D is the diagonal matrix of A.
2
Write P, the matrix of eigenvectors. For our example with two eigenvectors, P would be:
- ${\begin{bmatrix}-1&7\\1&8\end{bmatrix}}$
- Where the first column is the eigenvector of lambda = -5, and the second column is the eigenvector for lambda = 10.
3
Find the inverse of P. For our example, the inverse of P (P^-1) is:
- ${\begin{bmatrix}-8/15&7/15\\1/15&1/15\end{bmatrix}}$
- Check out our expert guide to finding the inverse of a 3x3 matrix if you need a refresher!
4
Solve for the diagonal matrix D. Use the equation P^-1 * A * P = D to diagonalize the matrix A. Continuing our example:
- ${\begin{bmatrix}-8/15&7/15\\1/15&1/15\end{bmatrix}}{\begin{bmatrix}2&7\\8&3\end{bmatrix}}{\begin{bmatrix}-1&1\\7&8\end{bmatrix}}=D$
- $D={\begin{bmatrix}-5&0\\0&10\end{bmatrix}}$
- You’re done! You have diagonalized a matrix. For more linear algebra info, check out how to divide matrices and how to transpose a matrix.