Symmetry of Orthogonally diagonalizable matrix

[DmytriE]

Can someone confirm or refute my thinking regarding the diagonalizability of an orthogonal matrix and whether it's symmetrical?

A = [b₁, b₂, ..., b_n] | H = Span {b₁, b₂, ..., b_n}. Based on the definition of the span, we can conclude that all of vectors within A are linearly independent. Furthermore, we can then conclude that Rank(A) = n.

If Rank(A) = n then none of the vectors in A can be made as a linear combination of the other n-1 vectors. Since A can be row-reduced to an identity matrix and the transpose of the identity matrix is itself. Can it be concluded that the original matrix A is also symmetrical (A^T = A)?

 

[Fredrik]

Based on the definition of the span, we can conclude that all of vectors within A are linearly independent.

This is incorrect. However, if A is orthogonal, then you can use that an orthogonal matrix is invertible (the transpose is the inverse), and that the columns of an invertible matrix are linearly independent.

 

[DmytriE]

This is incorrect. However, if A is orthogonal, then you can use that an orthogonal matrix is invertible (the transpose is the inverse), and that the columns of an invertible matrix are linearly independent.

When you say that this is incorrect, are you referring to my rational regarding symmetrical matrices or vector linear independence?

 

[Fredrik]

I was referring to the part of your post that I quoted. There's nothing about the definition of "span" that allows you to conclude that $\{b_1,\dots,b_n\}$ is linearly independent.

 

[micromass]

I was referring to the part of your post that I quoted. There's nothing about the definition of "span" that allows you to conclude that $\{b_1,\dots,b_n\}$ is linearly independent.

Unless of course you know somehow that the dimension of $\text{span}\{b_1,...,b_n\}$ is $n$.

 

[HallsofIvy]

Can someone confirm or refute my thinking regarding the diagonalizability of an orthogonal matrix and whether it's symmetrical?

A = [b₁, b₂, ..., b_n] | H = Span {b₁, b₂, ..., b_n}.

Do you intend here that H is itself n dimensional? If so, you need to say that. If not, your following statement is not true.

Based on the definition of the span, we can conclude that all of vectors within A are linearly independent. Furthermore, we can then conclude that Rank(A) = n.

If Rank(A) = n then none of the vectors in A can be made as a linear combination of the other n-1 vectors. Since A can be row-reduced to an identity matrix and the transpose of the identity matrix is itself. Can it be concluded that the original matrix A is also symmetrical (A^T = A)?

 

[Hawkeye18]

Orthogonal matrices are a particular case of unitary matrices, which are in turn a particular case of normal matrices. Any normal matrix $A$ is orthogonally diagonalizable, i.e. represented as $A=UDU^*$, where $U$ is a unitary and $D$ is diagonal matrix (generally with complex coefficients).

Orthogonal matrices do not need to be symmetric, a rotation matrix $$R_\alpha = \left(\begin{array}{cc} \cos \alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{array}\right)$$ with $\alpha\ne \pi n$ can serve as an example.

On the other hand, there are symmetric orthogonal matrices, for example the identity matrix $I$ or the matrices $$\left(\begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array}\right), \qquad \left(\begin{array}{cc} \cos \alpha & \sin\alpha \\ \sin\alpha &- \cos\alpha \end{array}\right) $$ (the last two matrices are unitarily equivalent).