Let A=[ 4 -1 -1 -1 4 -1 -1 -1 4 ] and 𝐯=[ 1 1 1 ] . Verify t

step 1 We are given a matrix A, vector V, and an eigenvalue of A. And we are asked to show that 5 is an

Let A=[ 4 -1 -1 -1 4 -1 -1 -1 4 ]...

Key Concepts

Orthogonal Diagonalization

Orthogonal diagonalization is the process of finding an orthogonal matrix that transforms a given square matrix into a diagonal matrix where the diagonal entries are the eigenvalues. In this process, the columns of the orthogonal matrix are normalized eigenvectors of the original matrix. This technique is particularly useful for symmetric matrices, where the spectral theorem guarantees the existence of such an orthogonal diagonalization.

Eigenvalues and Eigenvectors

An eigenvalue is a scalar associated with a square matrix that, when the matrix acts on its corresponding eigenvector, the output is the scalar multiple of that eigenvector. Specifically, for a matrix A and vector v, if Av = ?v, then ? is called the eigenvalue corresponding to eigenvector v. Verification of an eigenpair involves checking this equation, which indicates that the direction of the eigenvector is preserved under the transformation defined by the matrix.

Symmetric Matrix

A symmetric matrix is one that is equal to its transpose, meaning that A = A?. Such matrices have real eigenvalues and their eigenvectors can be chosen to be orthogonal. This property simplifies many matrix computations and is fundamental in applications involving quadratic forms and spectral analysis.

Spectral Theorem

The spectral theorem states that every symmetric matrix can be diagonalized by an orthogonal matrix, meaning there exists an orthonormal basis comprised entirely of eigenvectors of the matrix. This theorem underpins many theoretical and practical results in linear algebra, ensuring that symmetric matrices have real eigenvalues and a complete set of orthogonal eigenvectors.

Transcript

00:01 We are given a matrix a, vector v, and an eigenvalue of a.

00:08 And we are asked to show that 5 is an eigenvalue of a and v is an associated eigenvector.

00:19 And then we are asked to orthogonally diagonalize a.

00:27 Check to see if 5 is an eigenvalue of a.

00:32 We know that an eigenvalue satisfies the characteristic equation of a, is you want to check to see what the determinant of a minus 5i is.

00:52 This gives us the determinant of the matrix the same as a, except for now the diagonals are negative 1 instead of 4.

01:04 So negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1.

01:12 And we can see this determinant is clearly going to be 0, since we have duplicate 1, rows and therefore it follows that lambda is an eigenvalue value of a.

01:31 Check to see if v is an eigenvector in this eigen space.

01:36 We want to see if v satisfies the equation a minus 5i times v equals 0, or alternatively we could compute a times v to see if it's equal to 5i times v.

02:14 We have that a times v is equal to 4 -1 -1 -negative 1 -4 -negative 1.

02:25 So what do we have? we have 4 times 1 is 4 minus 1, minus 1, so 2.

02:40 Negative 1 times 1 is negative 1, 4 times 1 is negative 1.

02:46 Again 2.

02:50 And finally, negative 1 times 1 is negative 1, negative 1, negative 1, 4 times 1 is negative 1, 4 times 1 is 4, so again 2.

03:43 We have that this is equal to 2 times i times, instead of doing that, if we calculate a minus 5i v, we'll get something a little bit different.

04:20 So this equation implies that a minus 5i v is equal to, this is matrix of negative 1 times v, so we have negative 1 plus negative 1 plus negative 1, so negative 3 all the way down.

04:58 So it's clear that v is not an eigenvector actually associated with this eigenvalue.

05:09 But we have shown that av is equal to 2 -2, which is equal to 2 times v.

05:31 And so b is an eigenvector of a associated with an eigenvalue lambda equals 2.

06:00 Now notice that transpose of a is equal to, diagonal stays the same, 444, and then we exchange negative 1 with negative 1, negative 1 with negative 1, and negative 1 with negative 1.

06:16 We recognize this is the same as a, so we see see that a is a symmetric matrix since a transpose equals a, and therefore it follows that a is orthogonally diagonalizable.

06:48 To orthogonally diagonalize a, you define a complete set of orthonormal eigenvectors.

06:56 We have one eigenvector.

06:59 V is the eigenvector associated with the eigenvalue lambda 2.

07:15 We don't know, however, the multiplicity of eigenvalue lambda 2, which we could calculate.

07:31 We could also calculate the multiplicity of eigenvalue lambda 5, and use that to find the multiplicity of eigenvalue lambda 2.

07:37 So let's do that.

07:39 We have that a minus saying lambda 1 is equal to 5.

07:49 Igen value must satisfy a minus 5i, v1 equals 0.

07:58 And so we have that this coefficient matrix negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, negative 1, reduces to the matrix 1 -1 -1 -0 -0 -0 -0.

08:21 So we have 2 rows, which tells us that this eigenspace has dimension 2, and we obtain the system of equations, v11 plus v12 plus v13 equals 0.

08:33 And so we have that v13 is equal to negative v11 minus v12.

08:48 And if we take v111 to be the parameter s and v12 to be the parameter t, then the general form for the eigenvector v1 is s t negative s minus t, which can be rewritten as the linear combination s times 1 0, negative 1, plus t times 0, 1, negative 1.

09:18 So we obtain the eigenvector v1, if we take s to be 1 and t to be 0, which is 1, 0, negative 1, and we obtain the eigenvector v2.

09:29 If s is 0 and t is equal 1, which is the vector 0, 1, negative 1.

09:35 And by construction we have that v1 and v2 are linearly independent, and span, the eigenspan, so it follows that v1 and v2 are a basis for this eigenspace...

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