- Thread starter
- #1

- Thread starter dingo
- Start date

- Thread starter
- #1

- Feb 15, 2012

- 1,967

Basis for what?

- Thread starter
- #3

I mean a basis of vectors for the 2-dimensional vector spaceBasis for what?

- Feb 15, 2012

- 1,967

(4,7) is not a basis for $\Bbb R^2$, which having dimension 2, needs a basis of 2 vectors.

- Thread starter
- #5

Yes, you are right. My amended version is to get the change-of-basis matrix from basis A=(1,0),(4,7) to basis B=(0,2),(2,1)(4,7) is not a basis for $\Bbb R^2$, which having dimension 2, needs a basis of 2 vectors.

- Jan 30, 2012

- 2,528

- Jan 26, 2012

- 236

Say $(b_1,b_2)$ is an (ordered) basis for $\mathbb{R}^2$ and $(b_1',b_2')$ is another (ordered) basis for $\mathbb{R}^2$. We can express $b_1' = Ab_1 + Bb_2$ and $b_2' = Cb_1 + Db_2$. Then the change of basis matrix from $(b_1,b_2)$ to $(b_1',b_2')$ is given by,Hi

How can I get the change-of-basis matrix from basis (4,7) to basis (2,1)

Also, how do I use it once I get it.

$$ \begin{bmatrix} A & C \\ B & D \end{bmatrix} $$

- Admin
- #8

- Mar 5, 2012

- 9,416

Hi dingo!Yes, you are right. My amended version is to get the change-of-basis matrix from basis A=(1,0),(4,7) to basis B=(0,2),(2,1)

Suppose $(x_A,y_A)$ is a vector with respect to basis A and let $(x_B,y_B)$ be the same vector with respect to basis B.

Then with respect to the standard basis $\{(1,0), (0,1)\}$ they are both equal to:

$$\begin{bmatrix}1&4\\0&7\end{bmatrix} \begin{pmatrix}x_A\\y_A\end{pmatrix} = \begin{bmatrix}0&2\\2&1\end{bmatrix} \begin{pmatrix}x_B\\y_B\end{pmatrix}$$

Solve for $\begin{pmatrix}{x_B\\y_B}\end{pmatrix}$ to find the change-of-basis matrix.

- Jan 30, 2012

- 2,528

I think one has to solve for $(x_A,y_A)$ to find the change-of-basis matrix:Solve for $\begin{pmatrix}{x_B\\y_B}\end{pmatrix}$ to find the change-of-basis matrix.

\[

\begin{pmatrix}{x_A\\y_A}\end{pmatrix}=C \begin{pmatrix}{x_B\\y_B}\end{pmatrix}

\]

where $C$ is the required matrix.

(To the OP: this is a theorem, not the definition of a change-of-basis matrix.)

- Admin
- #10

- Mar 5, 2012

- 9,416

As I interpret it, the change-of-basis matrix from basis A to basis B would convert a vector with respect to basis A to a vector with respect to basis B.I think one has to solve for $(x_A,y_A)$ to find the change-of-basis matrix:

\[

\begin{pmatrix}{x_A\\y_A}\end{pmatrix}=C \begin{pmatrix}{x_B\\y_B}\end{pmatrix}

\]

where $C$ is the required matrix.

(To the OP: this is a theorem, not the definition of a change-of-basis matrix.)

So:

\[

\begin{pmatrix}{x_B\\y_B}\end{pmatrix}=C \begin{pmatrix}{x_A\\y_A}\end{pmatrix}

\]

But to be honest, I always get confused with what is from and what is to.

Edit: So I usually take a peek in a text book what they feel what is from and what is to, and then copy them.

- Jan 30, 2012

- 2,528

In contrast, the COB matrix $C$ can be considered as a matrix of a linear operator that maps $a_1,a_2$ into $b_1,b_2$, respectively. Then some vector $v$ with coordinates $(x_A,y_A)$ in basis $A$ gets mapped into a different vector $v'$ with some coordinates $(x_A',y_A')$

Thus, if $C$ is the COB matrix from $A$ to $B$ or, which is the same, the matrix of a linear operator that maps $A$ to $B$, then:

- If
*the same*vector $v$ has coordinates $(x_A,y_A)$ and $(x_B,y_B)$ in $A$ and $B$, respectively, then

\[

\begin{pmatrix}{x_A\\y_A}\end{pmatrix}= C\begin{pmatrix}{x_B\\y_B}\end{pmatrix}

\]

("old" coordinates are expressed through the "new" ones). - If a vector $v$ with coordinates $(x_A,y_A)$ in $A$ is mapped into a different $v'$ with coordinates $(x_A',y_A')$ still in $A$, then

\[

\begin{pmatrix}{x_A'\\y_A'}\end{pmatrix}= C\begin{pmatrix}{x_A\\y_A}\end{pmatrix}

\]

("new" coordinates are expressed through the "old" ones).

- Admin
- #12

- Mar 5, 2012

- 9,416

To figure this out and see if it is really as counter-intuitive as you write, I've googled "change of basis matrix". First hit was on wikipedia: https://en.wikipedia.org/wiki/Change_of_basis.

Unfortunately this article does not define a COB matrix sharply.

The best I can find is a section that is labeled

I'm afraid that other google hits do not really give any further enlightenment.

To be fair, whenever I try to help someone with this topic, I always refer to examples in their text book and copy whatever is there.

- Jan 30, 2012

- 2,528

- Jan 26, 2012

- 236

Who needs change-of-basis when one has commutative diagrams?

- Feb 15, 2012

- 1,967

To determine a vector by its coordinates, one needs a priori some idea of "what the coordinates mean". For example, if I tell you to "go that way 1, and turn left and go 4", a natural question to ask is: "which way" is "that" way (north? south? down a certain road?), and 1 and 4 "whats" (feet? miles? kilometers? blocks?)?.

So, given a basis $B = \{v_1,v_2\}$ of a two-dimensional vector space $V$ over a field $F$, I will write:

$[c_1,c_2]_B$ as shorthand for the formal linear combination:

$c_1v_1 + c_2v_2$.

It is important to realize that a matrix $A$ DOES NOT REPRESENT a linear transformation, per se. It rather represents a linear transformation GIVEN bases for the domain space, and co-domain space. If our linear transformation is $T$, we might write:

$A = [T]_B^C$

to denote that:

$[T]_B^C[v]_C = [T(v)]_B$

(It is a matter of choice which base one puts "up" or "down", I have made MY choice so that "up and down cancel" mimicking the Einstein summation convention).

When one writes an element of $\Bbb R^2$ as $(x,y)$, one is actually appealing to the "standard basis": $\{(1,0),(0,1)\}$ sometimes written as $\{e_1,e_2\}$ or $\{\mathbf{i},\mathbf{j}\}$. This basis is "invisible" because the coordinates in this basis match the vector itself.

But a vector space is not, "linear combinations of basis elements" per se, a vector space does not "care" which coordinates you use. A basis is a way of DESCRIBING a vector space, and different descriptions (different coordinate systems) are possible for the SAME vector space.

In $\Bbb R^2$ any pair of linearly independent vectors, can be used as a basis. In what follows I will call the bases like so:

$C = \{(1,0),(4,7)\} = \{u_1,u_2\}$

$B = \{(0,2),(2,1)\} = \{v_1,v_2\}$.

So, for example, $[1,0]_C = 1u_1 + 0u_2 = 1(1,0) + 0(4,7) = (1,0) + (0,0) = (1,0)$ whereas:

$[0,1]_C = 0u_1 + 1u_2 = 0(1,0) + 1(4,7) = (0,0) + (4,7) = (4,7)$

The "change-of-basis" matrix from $C$ to $B$ is simply the matrix:

$

It should be clear that the first column of this matrix will be the image of $[1,0]_C$, that is whatever (1,0) is in $B$-coordinates. To find this, we set:

$(1,0) = [a,b]_B = av_1 + bv_2 = a(0,2) + b(2,1) = (0,2a) + (2b,b) = (2b,2a+b)$

and solve for $a$ and $b$. This makes it clear that:

$b = \frac{1}{2}$

$a = -\frac{1}{4}$

Similarly, the second column of the matrix will be the image of $[0,1]_C$, that is, whatever (4,7) is in $B$-coordinates. Again, we solve:

$(4,7) = [a',b']_B = a'v_1 + b'v_2 = a'(0,2) + b'(2,1) = (2b',2a'+b')$ giving:

$b' = 2$

$a' = \frac{5}{2}$.

So our matrix should be:

$A = \begin{bmatrix}-\frac{1}{4}&\frac{5}{2}\\ \frac{1}{2}&2 \end{bmatrix}$

Let's, as a sanity check, verify that this matrix does what we want it to. Let's pick a random vector in $\Bbb R^2$, say (-2,5). If we've done our job right, then:

$A([v]_C) = [v]_B$

So first, we find what (-2,5) is in $C$-coordinates. This is the same process we've done twice already:

$(-2,5) = [r,s]_C = r(1,0) + s(4,7) = (r+4s,7s)$, yielding:

$r = -\frac{34}{7}$

$s = \frac{5}{7}$

Computing $A([-\frac{34}{7},\frac{5}{7}]_B)$ we obtain:

$[3,-1]_B$. Is this (-2,5) in $B$-coordinates? Let's see:

$[3,-1]_B = 3(0,2) - 1(2,1) = (0,6) - (2,1) = (-2,5)$.

Why, yes, it is.

*******************

There is an alternate way to do this, by relating both $B$ and $C$ to the standard basis. What we do here is find the change-of-basis matrices that do this:

$P: B \to \text{standard}$

$Q: C \to \text{standard}$

then it stands to reason that $P^{-1}$ is the matrix that does this:

$P^{-1}: \text{standard} \to B$

so that $P^{-1}Q$ does this:

$P^{-1}Q: C \to \text{standard} \to B$

which is what we want. The matrices $P$ and $Q$ are easy to write down, the columns are just the vectors of $B$ and $C$ in the standard basis. That is:

$P = \begin{bmatrix}0&2\\2&1 \end{bmatrix}, Q = \begin{bmatrix}1&4\\0&7 \end{bmatrix}$.

(Take a moment to see why this is so).

The difficulty here is in inverting $P$, but for a 2x2 matrix this is not so difficult, we find that:

$P^{-1} = -\frac{1}{4}\begin{bmatrix}1&-2\\-2&0 \end{bmatrix}$

and thus:

$P^{-1}Q = \begin{bmatrix}-\frac{1}{4}&\frac{1}{2}\\ \frac{1}{2}&0 \end{bmatrix}\begin{bmatrix}1&4\\0&7 \end{bmatrix}$

$= \begin{bmatrix}-\frac{1}{4}&\frac{5}{2}\\ \frac{1}{2}&2 \end{bmatrix}$

which is the same matrix we obtained above.

Perhaps it is just me, but I found most of the previous posts nearly impossible to decipher. The above approach was adapted largely from a similar discussion on Paul's Notes on Linear Algebra site, which unfortunately no longer exists.

- Jan 30, 2012

- 2,528

There is a type mismatch here. In order for the notation $[v]_C$ to make sense, $v$ has to be a pair of coordinates rather than a vector. But $T$ acts on vectors, not coordinates, so then $T(v)$ does not make sense. I assume that, contrary to the previous convention, $[v]_C$ denotes the coordinates of $v$ in the basis $C$.So, given a basis $B = \{v_1,v_2\}$ of a two-dimensional vector space $V$ over a field $F$, I will write:

$[c_1,c_2]_B$ as shorthand for the formal linear combination:

$c_1v_1 + c_2v_2$.

It is important to realize that a matrix $A$ DOES NOT REPRESENT a linear transformation, per se. It rather represents a linear transformation GIVEN bases for the domain space, and co-domain space. If our linear transformation is $T$, we might write:

$A = [T]_B^C$

to denote that:

$[T]_B^C[v]_C = [T(v)]_B$

- Feb 15, 2012

- 1,967

Yes, I am "abusing the notation" somewhat to write:There is a type mismatch here. In order for the notation $[v]_C$ to make sense, $v$ has to be a pair of coordinates rather than a vector. But $T$ acts on vectors, not coordinates, so then $T(v)$ does not make sense. I assume that, contrary to the previous convention, $[v]_C$ denotes the coordinates of $v$ in the basis $C$.

$[v]_C = [c_1,c_2]_C$ where:

$v = c_1u_1 + c_2u_2$

so that if $[T]_B^C = A$ we have:

$A((c_1,c_2)^T) = [d_1,d_2]_B$

where $[d_1,d_2]_B = d_1v_1 + d_2v_2 = T(v)$.

In other words, I am using "square brackets tagged with a subscript" to indicate which basis my coordinates express something in, and ALSO using them "conceptually" to indicate when I am writing a vector $v$ (which is "basis-less") in some particular basis.

The situation is complicated somewhat by the fact that we "identify" the representation of an element of $F^n$ with its coordinates in the standard basis, although these are actually two different things. So, in $\Bbb R^2$ for example, we will talk about "the vector (2,1)" which implicitly assumes a basis, whereas in an actual (physical) situation, we may have a force of a certain magnitude acting in a certain direction, and how we resolve that into coordinates depends essentially on where we put the origin, and how we orient our axes (these are choices we make to simplify calculation, and are not inherent in the situation).

A similar ambiguity results when we use permutation representations to represent permutation groups themselves.

Nonetheless, despite the ambiguity, the notation is useful. If you can see a "better" way to preserve rigor AND clarity, by all means, present it.