Relationship between eigenspace and column space

[ybhan23]

Is it true that if an n by n matrix A has n-linearly independent eigenvectors, then it must also be invertible because these n-eigenvectors span n-space. But does this reasoning work the other way around: that is if A is invertible, does that imply n-linearly independent eigenvectors can be found? More generally, is there some connection between the column space and eigenspace?

On a side note, here is an unrelated but interesting question: why is it that if the square of a matrix is the negative identity matrix (-I), then it implies that the matrix has an even number of columns and rows?

Thanks in advance

 

[morphism]

Is it true that if an n by n matrix A has n-linearly independent eigenvectors, then it must also be invertible because these n-eigenvectors span n-space.

What if 0 is an eigenvalue?

But does this reasoning work the other way around: that is if A is invertible, does that imply n-linearly independent eigenvectors can be found?

No. Consider $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

On a side note, here is an unrelated but interesting question: why is it that if the square of a matrix is the negative identity matrix (-I), then it implies that the matrix has an even number of columns and rows?

Thanks in advance

If A^2=-I then det(A^2)=det(-I). Try to see if you can use this to answer your question.

 

[ybhan23]

Thanks for your answer. The third question is completely clear to me now. But for the first and second questions, are there no connections between eigenspace and column space?

 

[ajkoer]

If A =P'EP where P'P = I and E is a diagonal matrix of eigenvalues, then:

A^-1 = P^-1* E^-1 * (P')^-1

But P^-1 = P' and (P')^-1 = P

So, A^-1 = P' * E^-1 * P

So as long as the eigenvalues are non-zero, P is the eigenvector for both A and A^-1.
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Now, the column space is the row space of the transpose. Assuming again that A=P'EP then:

A' = (P'EP)' = P'E'P = P'EP

so the dimension of eigenspace of the column space and row space are equal. In the event that E is upper diagonal form (not a diagonal matrix), I believe a similar statement can be made.

 

[blue_raver22]

.

The relationship between the eigenspace and column space of a matrix is an important concept in linear algebra. The eigenspace of a matrix is the set of all eigenvectors associated with a particular eigenvalue, while the column space is the span of the columns of the matrix.

It is true that if a square matrix A has n linearly independent eigenvectors, then it must also be invertible. This is because the n eigenvectors span the entire n-dimensional space, and therefore, any vector in the space can be expressed as a linear combination of these eigenvectors. This means that A can be diagonalized, with the eigenvectors as the columns of the diagonalizing matrix, and the eigenvalues along the diagonal. Since A is diagonalizable, it must also be invertible.

However, the reverse is not necessarily true. If a matrix A is invertible, it does not necessarily imply that n linearly independent eigenvectors can be found. For example, a matrix with all distinct eigenvalues will have n linearly independent eigenvectors, but a matrix with repeated eigenvalues may not.

There is a connection between the column space and eigenspace of a matrix. The eigenspace of a matrix is a subspace of the column space, and the column space can be thought of as the span of the eigenvectors. This means that the eigenvectors of a matrix are also part of the column space, and the column space can be decomposed into eigenspaces corresponding to different eigenvalues.

As for the unrelated question, if the square of a matrix is the negative identity matrix (-I), then it implies that the matrix has an even number of columns and rows because the determinant of a matrix is equal to the product of its eigenvalues. Since the square of a matrix has eigenvalues that are the squares of the original matrix's eigenvalues, the determinant will be the square of the original determinant. Therefore, if the determinant is -1, the square of the determinant will be 1, and the original determinant must have been -1 or 1. Since the determinant of a matrix with an odd number of rows and columns must be an odd number, it cannot be equal to -1, and therefore, the matrix must have an even number of rows and columns.