Let b be a nonzero vector (from R^n). Then, bb' is an n x n matrix.
What are its eigenvectors and eigenvalues?
From linear algebra, we know that bb' is a rank one matrix. So it has one non-zero eigenvalue. The other n - 1 eigenvalues are all 0.
Now that is the non-zero eigenvalue? Is it positive or negative? What is the associated eigenvector?
What about the eigenvectors corresponding to the zero eigenvalue?
Recall that the eigenvector of a matrix M is defined as any vector x such that
M x = p x
where p is the eigenvalue corresponding to x.
Since the matrix was formed with b alone, one would guess that the eigenvector and eigenvalue is somehow related to b. Let us try b.
Now,
(bb') b = b (b'b) = (b'b) b
We use associativity of matrix multiplication above. In the last step, we could move the b'b to the front since it is a real number.
Thus b is indeed the eigenvector we are looking for with the eigenvalue b'b. Note that b'b is the length square of the vector b. Thus it is non-negative.
What about the other eigenvectors with 0 eigenvalues?
Let c be a vector from the orthogonal complement of b. Then, b'c = 0.
Thus, (bb') c = b (b'c) = b 0 = 0 = 0 c
Thus any vector orthogonal to b is an eigenvector of bb' with eigenvalue 0.
As we know from linear algebra, the dimension of the orthongonal complement of b is n - 1, so we can find n-1 linearly independent vectors c such that (bb') c = 0.
Thus we have all the eigenvalues and eigenvectors of bb'.
Tailpiece:
As a next step, let b and c be two vectors from R^n. Form the matrix
M = bb' + cc'
1. Is this always a rank 2 matrix?
2. What are the eigenvalues and eigenvectors of M in terms of b and c?
What are its eigenvectors and eigenvalues?
From linear algebra, we know that bb' is a rank one matrix. So it has one non-zero eigenvalue. The other n - 1 eigenvalues are all 0.
Now that is the non-zero eigenvalue? Is it positive or negative? What is the associated eigenvector?
What about the eigenvectors corresponding to the zero eigenvalue?
Recall that the eigenvector of a matrix M is defined as any vector x such that
M x = p x
where p is the eigenvalue corresponding to x.
Since the matrix was formed with b alone, one would guess that the eigenvector and eigenvalue is somehow related to b. Let us try b.
Now,
(bb') b = b (b'b) = (b'b) b
We use associativity of matrix multiplication above. In the last step, we could move the b'b to the front since it is a real number.
Thus b is indeed the eigenvector we are looking for with the eigenvalue b'b. Note that b'b is the length square of the vector b. Thus it is non-negative.
What about the other eigenvectors with 0 eigenvalues?
Let c be a vector from the orthogonal complement of b. Then, b'c = 0.
Thus, (bb') c = b (b'c) = b 0 = 0 = 0 c
Thus any vector orthogonal to b is an eigenvector of bb' with eigenvalue 0.
As we know from linear algebra, the dimension of the orthongonal complement of b is n - 1, so we can find n-1 linearly independent vectors c such that (bb') c = 0.
Thus we have all the eigenvalues and eigenvectors of bb'.
Tailpiece:
As a next step, let b and c be two vectors from R^n. Form the matrix
M = bb' + cc'
1. Is this always a rank 2 matrix?
2. What are the eigenvalues and eigenvectors of M in terms of b and c?
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