Proving set is orthonormal

Question

Let $B = \{\vec{v_1} ... \vec{v_n}\}$ be basis for an n-dimensional vector space V. If $B[\vec{v_i}]_B = \vec{v_i}$ for all $1 \le i \le k$ Does that mean that $B$ is an orthogonal set? If, yes what is the proof or theorem to use for a proof? Also, does that mean that $B$ is a standard basis?

Answer 1

It does not seem that $$ B[\vec{v}_i]_B=\vec{v}_i $$ indicates that the collection of vectors are orthonormal. In fact, this statement appears to be always true. Let's break it down:

• The term $$ [\vec{v}_i]_B $$ means that we're writing $\vec{v}_i$ in terms of the basis $B$ and collecting the coefficients. Since $\vec{v}_i$ is in the basis, we see that $$ 0\vec{v}_1+0\vec{v}_2+\dots+0\vec{v}_{i-1}+1\vec{v}_i+0\vec{v}_{i+1}+\dots+0\vec{v}_n=\vec{v}_i. $$ What $[\vec{v}_i]_B$ means is that you take the coefficients of the basis vectors of $B$ and put them into a vector. Therefore, $$ [\vec{v}_i]_B=\begin{bmatrix}0\\0\\\vdots\\0\\1\\0\\\vdots\\0\end{bmatrix}, $$
  where the $1$ is in the $i^{\text{th}}$ position. Therefore $$ [\vec{v}_i]_B=\vec{e}_i $$ where $\vec{e}_i$ is the $i^{\text{th}}$ standard basis vector (the standard basis is the one where the basis vectors have all zeros except for one $1$ - in other words, the columns of the identity matrix).

• Now, let's calculate $$ B[\vec{v}_i]_B. $$ One way to do matrix vector multiplication is to look at a product $B[\vec{v}_i]_B$ as a weighted sum of the columns of $B$. In other words, $$ B[\vec{v}_i]_B=B\vec{e}_i=0B_1+0B_2+\dots+0B_{i-1}+1B_i+0B_{i+1}+\dots+0B_n. $$ Since the columns of $B$ are the basis vectors, it follows that $$ B[\vec{v}_i]_B=0\vec{v}_1+0\vec{v}_2+\dots+0\vec{v}_{i-1}+1\vec{v}_i+0\vec{v}_{i+1}+\dots+0\vec{v}_n=\vec{v}_i. $$

Therefore, the given equality always holds - we never had to use orthogonality - so it really has very little to do with orthogonality.