Let $B_1 = \{v_1, v_2, v_3\} $ be a basis for $\mathbb R^3$

Let $B_2 = \{o_1, o_2, o_3\}$ be an orthonormal basis after executing the Gram-Schmidt algorithm on $B_1$

Let $$ P = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 3 & 1 \\ 2 & 0 & 0 \\ \end{pmatrix} $$

Can $P$ be the change of basis matrix from $B_1$ to $B_2$?

It feels a bit tricky and i was not able to figure out how to approach this question... any help much appreciated!

  • $\begingroup$ You need it to not cause input to lose any dimensions. I.e. null space must be empty. In other words, rank must be full. $\endgroup$ – mathreadler Jun 20 '18 at 13:02

Yes, this is possible. Firstly notice that $P$ is invertible, so the columns are linearly independent. This is necessary. Let us try to construct an example of where this is the case. Using the standard `change of base diagram' with the identity transformation we can obtain the following formula, for all vectors $x \in \mathbb{R}^3$: $$ Id_{B_1}^{B_2} \cdot co_{B_1}(x) = co_{B_2}(x). $$ If we now use $x = v_1, x=v_2, x=v_3$ respectively, and using the matrix $P = Id_{B_1}^{B_2}$ that you provided, we obtain: \begin{align*} Id_{B_1}^{B_2} \cdot e_1 = co_{B_2}(v_1) = [1,2,2]^\top \\ Id_{B_1}^{B_2} \cdot e_2 = co_{B_2}(v_2) = [0,3,0]^\top \\ Id_{B_1}^{B_2} \cdot e_3 = co_{B_2}(v_3) = [1,1,0]^\top \\ \end{align*} Thus, this gives $$ \begin{cases} v_1 = o_1 + 2o_2 + 2o_3 \\ v_2 = 3o_2 \\ v_3 = o_1+o_2. \end{cases} $$ Now pick any orthonormal basis, for example pick $\{o_1,o_2,o_3\} = \{e_1,e_2,e_3\}$. Then according to the formulas we obtain $$ v_1 = [1,2,2]^\top, v_2 = [0,3,0]^\top, v_3 = [1,1,0]^\top. $$ You can check that this is a basis of $\mathbb{R}^3$ and that the matrix for change of base from $\{v_1,v_2,v_3\}$ to $\{o_1,o_2,o_3\}$ is given by $P$. As an example consider the vector $$ x = [3,14,4]^\top $$ This has $co_{B_1}(x) = [2,3,1]^\top$. Now we multiply with $P$, this indeed gives $$ P \cdot co_{B_1}(x) = [3,14,4]^\top = co_{B_2}(x) $$


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