Change of basis matrix from standard basis to orthonormal basis is an orthogonal matrix

Question

Let $\alpha = \{v_1, ..., v_n \}$ be a orthonormal basis. Let $Q$ be the change of basis matrix from the standard basis to $\alpha$. We know that $$Q = \begin{bmatrix} | & & | \\ v_1 & \dots & v_n \\ | & & | \end{bmatrix}$$ And $$Q^t = \begin{bmatrix} - & v_1 & - \\ & \vdots & \\ - & v_n & - \end{bmatrix}$$ Then $Q^tQ$ clearly has entries $<v_i, v_j>$ So it is the identity. This is easy but how do I verify that $QQ^t$ also is the identity. I am having trouble because i have to take the rows of $Q$ and columns of $Q^t$. I tried showing that the rows of $Q$ form an orthonormal basis, but couldnt, is there any other way to show this? My textbook says that it is a similar computation to $Q^tQ$ but I cant see this.

Answer 1

The usual approach to showing that $QQ^T = I$ is, as I suggest in my comment, to note that $AB = I$ for square matrices $A,B$ generally implies that $BA = I$, perhaps using the fact that $B = A^{-1}$. In short, $$ Q^TQ = I \iff Q^{-1} = Q^T \iff QQ^T = I. $$ However, there is an approach here that is (arguably) more direct. Contrary to your textbook's claim, I would not call this a "similar computation to $Q^TQ$"; it is not generally true that a matrix having orthonormal columns should imply that it has orthonormal rows, and the fact that this changes in the case that the matrix is square requires justification.

Recall that for an orthonormal set $S = \{v_1,\dots,v_n\}$ and any vector $x$, the sum $$ \langle x,v_1\rangle v_1 + \cdots + \langle x,v_n\rangle v_n $$ denotes the projection of $x$ onto the span of $S$. If $S$ is a basis of $\Bbb R^n$, it follows that $$ x = \langle x,v_1\rangle v_1 + \cdots + \langle x,v_n\rangle v_n. $$ Thus, for every vector $x$, we have $$ (QQ^T)x = Q(Q^Tx) = \pmatrix{v_1 & \cdots & v_n} \pmatrix{\langle v_1,x \rangle\\ \vdots \\ \langle v_n,x \rangle} = \langle x,v_1\rangle v_1 + \cdots + \langle x,v_n\rangle v_n = x. $$ That is, we have $(QQ^T)x = x$ for all vectors $x \in \Bbb R^n$, from which it follows that $QQ^T = I$.