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Finding the Eigenvalues and Eigenvectors of a Matrix Finding the eigenvalues and eigenvectors of a matrix are fundamental operations in linear algebra. The Mathcad functions eigenvals and eigenvecs perform these operations. You can also use eigenvecs to determine whether a matrix can be diagonalized.

Eigenvalues and Eigenvectors A number c is an eigenvalue of a matrix A if there is a non-zero vector v, called an eigenvector for c, so that the following equation is true:

A v = c v Here's an example:

9 2 5 2  2   A   2 5 2  5 1 2   2 2  1  v   1  1    Since

 5  A  v   5  5   

is equal to

 5  5 v   5  5   

v is an eigenvector for A, corresponding to eigenvalue c = 5. Geometrically, this means that multiplying A times v stretches v by a factor of 5.

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The function eigenvals returns a vector containing the eigenvalues of A.

 1  eigenvals ( A)   5  4    The function eigenvecs returns a matrix whose columns are the corresponding eigenvectors of A.

U  eigenvecs ( A)

 0.267 0.577 0.196  U   0.535 0.577 0.784   0.802 0.577 0.588    Note that the second column of U is a scalar multiple of the eigenvector v above. To return a single column of U, first type

ORIGIN  1 Doing so starts the numbering of the columns at 1. Next, type U followed by [Ctrl] 6 to insert the column operator.

 U Finally, type the number of the column in the placeholder to the right of U. For example, the third column of U is

 0.196  3  U   0.784   0.588   

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You can verify that this is the eigenvector corresponding to 4, the third entry of the vector of eigenvalues.

 0.784   0.784  3  3  A U   3.138  is equal to 4 U   3.138   2.353   

 2.353   

Diagonalizing a Matrix A square matrix is called a diagonal matrix if its only non-zero entries are on the main diagonal. For example:

 1 0 0  0 5 0  0 0 4    The eigenvalues of a diagonal matrix are its diagonal entries. The corresponding eigenvectors are the standard basis vectors of R3. If a matrix A is not diagonal, it is often necessary to diagonalize it to find an invertible matrix U and a diagonal matrix D that satisfy the matrix equation

U

1

 A U = D

If you can find the matrices U and D, then A is diagonalizable. How can you tell if an n-by-n matrix A is diagonalizable? The answer is that A is diagonalizable if it has n linearly independent eigenvectors. You can determine whether this is true by evaluating eigenvecs(A) symbolically, provided that the entries of A are written as common fractions, such as

9 , rather than as decimals, 2

such as 4.5. If the result is a square matrix, then A is diagonalizable. Here's how to test whether a matrix is diagonalizable:

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Diagonalizability Test 1. Type

U  eigenvecs ( A) 2. Press [Ctrl] [.], followed by [Enter], to evaluate U with the symbolic equal sign.

· If U is a square matrix - it has the same number of columns as rows - then A is diagonalizable. U is invertible and satisfies the matrix equation above, where D is the diagonal matrix whose diagonal entries are the eigenvalues of A.

· If U has fewer columns than rows, A is not diagonalizable. Since U is not a square matrix, it cannot be invertible. Note: You must use the symbolic equal sign for this test, and the entries of A must be symbolic expressions. Evaluating U with the numerical equal sign always returns a square matrix, even when A is not diagonalizable. Let's apply the test to the matrix A in the example above:

U  eigenvecs ( A)

 1  3  U  2   3   1

1 3



1 



4   1  3 1



1 

Since U is a square matrix, A is diagonalizable. The diagonal matrix D is

 1 0 0  1 U  A U   0 5 0  0 0 4    The result is a diagonal matrix whose diagonal entries are the eigenvalues of A. Note: When you evaluate U symbolically, the columns are displayed in a different order than when you evaluate it numerically. However, this does not affect the test.

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Not all matrices are diagonalizable. Here's an example.

 2 0 1  B   1 2 2  0 0 3    If you apply the diagonalizability test to B, the result is

U  eigenvecs ( B)

 0 1  U  1 3  0 1    Since U is not a square matrix, B is not diagonalizable. What this means is that B does not have three linearly independent eigenvectors, so its eigenvectors do not span all of R3.

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