# Spectral Decomposition

Theorem 1 (Spectral Decomposition): Let A be a symmetric n × n matrix, then A has a spectral decomposition A = CDCT where C is a n × n matrix whose columns are unit eigenvectors C1, …, Cn corresponding to the eigenvalues λ1, …, λn of A and D is then × n diagonal matrix whose main diagonal consists of λ1, …, λn.

Proof: We prove that every symmetric n × n matrix is orthogonally diagonalizable by induction on n. The property is clearly true for n = 1. We assume that it is true for any n × n symmetric matrix and show that it is true for an n+1 × n+1 symmetric matrix A.

Let λ be any eigenvalue of A (we know by Property 1 of Symmetric Matrices that A has + 1 real eigenvalues) and let X be a unit eigenvector corresponding to λ. Thus AX = λX, and so XTAX = XTλX = λ(XTX) = λ(X ∙ X) = λ, showing that λ = XTAX.

By Property 3 of Linear Independent Vectors, we can construct a basis for the set of all n+1 × 1 column vectors which includes X, and so using Theorem 1 of Orthogonal Vectors and Matrices (Gram-Schmidt), we can construct an orthonormal basis for the set of n+1 × 1 column vectors which includes X. Now define B to be the matrix whose columns are the vectors in this basis excluding X. By Property 4 of Orthogonal Vectors and Matrices, B is an n+1 × n orthogonal matrix.

Note that (BTAB)T = BTATBT = BTAB since A is symmetric. This shows that BTAB is a symmetric n × n matrix, and so by the induction hypothesis there is an n × n diagonal matrix E whose main diagonal consists of the eigenvalues of BTAB and an orthogonal n × n matrix P such BTAB = PEPT. Now define the n+1 × n matrix Q = BP. Note that by Property 5 of Orthogonal Vectors and Matrices Q is orthogonal.

Now define the n+1 × n+1 matrix C whose first row is X and whose remaining rows are those of Q, i.e. C = [X, Q]. We now show that C is orthogonal. First we note that since X is a unit vector, XTX = X ∙ X = 1. Since the columns of B along with X are orthogonal, XTBj= X ∙ Bj = 0 for any column Bj in B, and so XTB = 0, as well as BTX = (XTB)T = 0. Finally since Q is orthogonal, QTQ = I.

Now

This completes the proof that C is orthogonal. We next show that QTAQ = E.

Next we need to show that QTAX = XTAQ = 0. Since

since A is symmetric, it is sufficient to show that QTAX = 0.

As we saw above, BTX = 0. Also, since λ is an eigenvalue corresponding to X, AX = λX. Thus,

Now

Defining

it follows that CTAC = D, and so

Note that at each stage of the induction, the next item on the main diagonal matrix of D is an eigenvalue of A and the next column in C is the corresponding eigenvector and that this eigenvector is orthogonal to all the other columns in C.

Observation: The spectral decomposition can also be expressed as A = $\sum_{j=1}^n \lambda_j C_j C_j^T$.

Example 1: Find the spectral decomposition of the matrix A in range A4:C6 of Figure 1.

Figure 1 – Spectral Decomposition

We calculate the eigenvalues/vectors of A (range E4:G7) using the supplemental function eVECTORS(A4:C6). Matrix C (range E10:G12) consists of the eigenvectors of A and matrix D (range I10:K12) consists of the square roots of the eigenvalues. You can check that A = CDCT using the array formula

=MMULT(E10:G12,MMULT(I10:K12,M10:O12))

Real Statistics Function: The Real Statistics Resource Pack provides the following function:

SPECTRAL(R1, iter): returns a 2n × n range whose top half is the matrix C and whose lower half is the matrix D in the spectral decomposition of CDCT of where A is the matrix of values in range R1.

Here iter is the number of iterations in the algorithm used to compute the spectral decomposition (default 100).

Real Statistics Data Analysis Tool: The Spectral Factorization option of the Real Statistics Matrix Operations data analysis tool also provides the means to output the spectral decomposition of a symmetric matrix.

Observation: As we have mentioned previously, for an n × n matrix A, det(A – λI) is an nth degree polynomial of form (-1)n $\prod_{i=1}^n {}$ (x – λi) where λ1, …., λn are the eigenvalues of A. If all the eigenvalues are distinct then we have a simpler proof for Theorem 1 (see Property 4 of Symmetric Matrices). But as we observed in Symmetric Matrices, not all symmetric matrices have distinct eigenvalues. This motivates the following definition

Definition 1: The (algebraic) multiplicity of an eigenvalue is the number of times that eigenvalue appears in the factorization (-1)n $\prod_{i=1}^n {}$ (x – λi) of det(A – λI).

Property 1: For any eigenvalue λ of a square matrix, the number of independent eigenvectors corresponding to λ is at most the multiplicity of λ.

Proof: Suppose λ1 is an eigenvalue of the n × n matrix A and that B1, …, Bk are k independent eigenvectors corresponding to λ1. By Property 3 of Linear Independent Vectors, there are vectors Bk+1, …, Bn such that B1, …, Bn is a basis for the set of n × 1 vectors. Now let B be the n × n matrix whose columns are B1, …, Bn. Since B1, …, Bn are independent, rank(B) = n and so B is invertible. By Property 9 of Eigenvalues and Eigenvectors we know that B-1AB and A have the same eigenvalues, and in fact they have the same characteristic polynomial.

Now consider AB. The first k columns take the form AB1, …, ABk, but since B1, …, Bk are eigenvectors corresponding to λ1, the first k columns are λB1, …, λBk. It now follows that the first k columns of B1AB consists of the vectors of the form D1, …, Dk where Dj consists of λ1 in row j and zeros elsewhere. This means that the characteristic polynomial of B1AB has a factor of at least (λ – λ1)k, i.e. the multiplicity of B1AB, and therefore A, is at least k.

Property 2: For each eigenvalue λ of a symmetric matrix there are k independent (real) eigenvectors where k equals the multiplicity of λ, and there are no more than k such eigenvectors.

Proof: By Theorem 1, any symmetric n × n matrix A has n orthonormal eigenvectors corresponding to its n eigenvalues. By Property 2 of Orthogonal Vectors and Matrices, these eigenvectors are independent. This shows that the number of independent eigenvectors corresponding to λ is at least equal to the multiplicity of λ. But by Property 5 of Symmetric Matrices, it can’t be greater than the multiplicity of λ, and so we conclude that it is equal to the multiplicity of λ.

By Property 1 of Symmetric Matrices, all the eigenvalues are real and so we can assume that all the eigenvectors are real too.

### 5 Responses to Spectral Decomposition

1. Naeem ullah says:

if 2 by 2 matrix is solved to find eigen value it will give one value it possible

• Charles says:

Sorry Naeem, but I don’t understand your comment. Are you looking for one value only or are you only getting one value instead of two?
Charles

2. Tapan kumar mahanta says:

Thanks a lot sir for your help regarding my problem. Keep it up sir.
You are doing a great job sir.
With regards
Tapan

3. Tapan kumar mahanta says:

For spectral decomposition As given at Figure 1
By taking the A matrix=[4 2 -1
2 3 1
-1 1 9],
when i am trying to find Eigen value and corresponding Eigen Vector by using eVECTORS(A).
I am only getting only one Eigen value 9.259961. How to get the three Eigen value and Eigen Vectors.

• Charles says:

You need to highlight the range E4:G7 insert the formula =eVECTORS(A4:C6) and then press Ctrl-Shift-Enter. Since eVECTORS is an array function you need to press Ctrl-Shift-Enter and not simply Enter.
Charles