Row space and column space of square matrix $A$ are the same. What does that mean?
I guess that means $A$ is symmetric or eigenvalues of $A$ are all different.
Are there other possibilities? What is the theorem for this?
Thanks.
Square matrix with identical row space and column space
linear algebra
Related Solutions
Recall that two vectors are orthogonal if and only if their inner product is zero. You are incorrect in asserting that if the columns of $Q$ are orthogonal to each other then $QQ^T = I$; this follows if the columns of $Q$ form an orthonormal set (basis for $\mathbb{R}^n$); orthogonality is not sufficient. Note that "$Q$ is an orthogonal matrix" is not equivalent to "the columns of $Q$ are pairwise orthogonal".
With that clarification, the answer is that if you only ask that the columns be pairwise orthogonal, then the rows need not be pairwise orthogonal. For example, take $$A = \left(\begin{array}{ccc}1& 0 & 0\\0& 0 & 1\\1 & 0 & 0\end{array}\right).$$ The columns are orthogonal to each other: the middle column is orthogonal to everything (being the zero vector), and the first and third columns are orthogonal. However, the rows are not orthogonal, since the first and third rows are equal and nonzero.
On the other hand, if you require that the columns of $Q$ be an orthonormal set (pairwise orthogonal, and the inner product of each column with itself equals $1$), then it does follow: precisely as you argue. That condition is equivalent to "the matrix is orthogonal", and since $I = Q^TQ = QQ^T$ and $(Q^T)^T = Q$, it follows that if $Q$ is orthogonal then so is $Q^T$, hence the columns of $Q^T$ (i.e., the rows of $Q$) form an orthonormal set as well.
There are several basis you can choose for a vector space.
Say $M$ is your matrix. Then $M\,\mathbb R^4$ is a vector space and since $\det(M)\neq 0$ it has dimension 4, that is any of its bases has four vectors.
Notice that any $v$ in this space can be written as $v=Mw$ for some $w$ so $v$ is a linear combination of the columns of the matrix, which means that the four vector columns of $M$ are a basis for the space.
Now say $\hat{M}$ is $M$ is in reduced column echelon form. Then the four vector columns of $\hat{M}$ are also a basis for the space.
We also have that $\mathbb{R}^4\, M$ is a vector space of dimension 4 since again the determinant of $M$ is not zero. Look at the space as the set of vectors $v$ such that $v=w \,M$ for some $w\in\mathbb{R}^4$. Any such $v$ is a linear combination of the row vectors of $M$. The same comments I made above with the respect the column echelon form also applies here with the row echelon form. The four row vectors of the row echelon matrix will be a basis for $\mathbb{R}^4\, M$.
Best Answer
It doesn't imply $A$ symmetric nor that all eigenvalues different. Easy to see when $A$ is invertible: then the row and column spaces are equal. But there are scores of non-symmetric invertible matrices, and you can easily choose them with repeated eigenvalues. For an extreme example of non-different eigenvalues, consider the identity $I_n$.
With the idea above you can get examples for all dimensions of the column/row space by putting a matrix as above in the upper left corner.