Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am taking a linear algebra class currently and working through Hoffman's textbook. One of the exercises I am unsure about is,

Find the coordinate matrix of the vector $\alpha=(1,0,1)$ in the basis of $\mathbb{C}^3$ consisting of the vectors $(2i,1,0),(2,-1,1),(0,1+i,1-i)$, in that order.

As I understand it, we want to write the vector $\alpha=(1,0,1)$ that is currently in terms of the standard basis $\mathbb{B}=\{(1,0,0),(0,1,0),(0,0,1)\}$, in terms of a new basis $\mathbb{B}'=\{(2i,1,0),(2,-1,1),(0,1+i,1-i)\}$. That is, we want to determine what matrix $P$ will satisfy $[\alpha]_{\mathbb{B}'}=P[\alpha]_{\mathbb{B}}$. There aren't many examples in Hoffman's textbook for actual computation, and what I recall from the lecture on the change of basis, we write $\mathbb{B}'$ in columns as,

$$Q= \left[ \begin{array}{ccc} 2i & 2 & 0 \\ 1 & -1 & 1+i \\ 0 & 1 & 1-i \\ \end{array} \right]$$

By inverting this matrix $Q$ we find that,

$$Q^{-1}= \left[ \begin{array}{ccc} \frac{1-i}{2} & -i & -1 \\ \frac{-i}{2} & -1 & i \\ \frac{i-1}{4} & \frac{1+i}{2} & 1 \\ \end{array} \right]$$

So, we then have that the coordinates $(x_{1}',x_{2}',x_{3}')$ of the vector $\alpha = (x_{1},x_{2},x_{3})$ in terms of the basis $\mathbb{B}'$ is given by,

$$\left[ \begin{array}{c} x_{1}' \\ x_{2}' \\ x_{3}' \\ \end{array} \right] = \left[ \begin{array}{ccc} \frac{1-i}{2} & -i & -1 \\ \frac{-i}{2} & -1 & i \\ \frac{i-1}{4} & \frac{1+i}{2} & 1 \\ \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ \end{array} \right]$$

So, we can then substitute $\alpha=(1,0,1)$ for $x_{1}, x_{2}, x_{3}$ to find the coordinates $x_{1}',x_{2}',x_{3}'$ in terms of the basis $\mathbb{B}'$. That is $P=Q^{-1}$. Substituting in these values we receive $\alpha'=(\frac{-1-i}{2},\frac{i}{2},\frac{3+i}{4})$.

Have I done this correctly? Even if I have done this correctly I am unsure why these procedure for determining the matrix $P$ such that $[\alpha]_{\mathbb{B}'}=P[\alpha]_{\mathbb{B}}$ works. Is there any way I can understand this more intuitively than memorizing a procedure for changing a basis (assuming I have done this correctly, if I haven't, please explain).

Thank you.

share|cite|improve this question
The simple way to check if you've done it correctly is to use the coordinates of $\alpha'$ as coefficients for a linear combination of vectors of $\mathbb{B}'$ and verify that you get the original vector back. That is, verify that $$\frac{-1-i}{2}(2i,1,0) + \frac{i}{2}(2,-1,1) + \frac{3+i}{4}(0,1+i,1-i) = (1,0,1).$$If this is the case, then $\alpha'$ is correct. If this is not the case, then you did something wrong. – Arturo Magidin Feb 12 '12 at 4:20
@ArturoMagidin, I checked through the computation you suggested and my answer was correct! Could you give a quick explanation as to why this procedure works or some motivation behind it? I couldn't find a good explanation about where the idea came from or why we would write, say, the basis vectors in terms of a column instead of rows, or we invert the matrix, etc. I'm at a real lack of understanding for the motivation behind these computations although I understand the theory in principle. – Samuel Reid Feb 12 '12 at 4:36
By definition, the coordinate vector of $\mathbf{v}$ relative to the ordered basis $\mathbb{B}=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$ is the vector $(\alpha_1,\ldots,\alpha_n)$ with the (unique) scalars such that $$\mathbf{v}=\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n.$$Every vector can be written uniquely in terms of the basis, and the coordinate vectors tell you how they are written in terms of the basis. – Arturo Magidin Feb 12 '12 at 5:01
@SamuelReid All it boils down to is solving a $3 \times 3$ linear system. This is all a matter of coming from definition!! – user38268 Feb 12 '12 at 11:44
up vote 2 down vote accepted

(I'll do it for dimension $n$ because the difficulty is the same)

Suppose you write $\{e_1,\ldots,e_n\}$ for the canonical basis (could be any basis, actually), and let $\{f_1,\ldots,f_n\}$ be another basis. You take a vector $x=\sum_{j=1}^n x_j\, e_j$ and you want to write it in the other basis.

You are given the vectors $\{f_1,\ldots,f_n\}$ in terms of the canonical basis, which means you have $\{p_{kj}\}$ such that $$ f_k=\sum_{j=1}^n p_{jk}e_j,\ \ \ k=1,\ldots,n. $$ Here you can think of $P=(p_{kj})$ as the matrix that has the coefficients of the $f_j$ in its columns. In a similar way we have coefficients $\{q_{jh}\}$ such that $$ e_j=\sum_{h=1}^nq_{hj}f_h\ \ \ j=1\,\ldots,n. $$ Combining the two expressions we get $$ f_k=\sum_{j=1}^n\sum_{h=1}^np_{jk}q_{hj}f_h=\sum_{h=1}^n\sum_{j=1}^nq_{hj}p_{jk}f_h =\sum_{h=1}^n(QP)_{kh}f_h. $$ By the uniqueness of the coefficients of a vector in a basis we get that $(QP)_{kh}$ is $1$ when $k=h$ and $0$ otherwise, i.e. $QP=I$. So $Q$ is the inverse matrix of $P$.

Now $$ x=\sum_{j=1}^n x_j\, e_j=\sum_{j=1}^n x_j\,\sum_{h=1}^n q_{hj}f_h =\sum_{h=1}^n \sum_{j=1}^n q_{hj}x_jf_h =\sum_{h=1}^n (QX)_h f_h. $$ In other words, the coefficients of the vector $x$ in the basis $\{f_1,\ldots,f_n\}$ are given by $P^{-1}X$, where $P$ is the matrix with the entries of the $f_k$ in its column, and $X$ are the entries of $x$ in the canonical basis.

share|cite|improve this answer
Excellent explanation, this is exactly what I was looking for! Thank you very much for the response. – Samuel Reid Feb 12 '12 at 5:25
You are welcome! – Martin Argerami Feb 12 '12 at 5:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.