Change of basis formula - intuition/is this true?

Question

Let $E$ and $F$ be two bases of the same $n$-dimensional vector space $U$.

Does the change of basis matrix from $E$ to $F$ have size $n \times n$? Give a counterexample or brief justification.

My thoughts:

The change of basis matrix from $E$ to $F$ is formed by expressing the coordinates of vector in $U$ with respect to $E$ as a coordinate with respect to $F$ (Although, I'd also like someone to help me make this statement clearer).

Since $E$ and $F$ both have dimension $n$, this matrix will have $n$ columns (as there are $n$ basis vectors). However, I'm not sure if it'll need $n$ rows.

Any help with this would be much appreciated!

tell me if this can help https://math.stackexchange.com/questions/1092328/relationship-between-t-mathcalb-and-t-mathcalb/1092384#1092384 — janmarqz, May 19 '17 at 12:49
Maybe it help if you just look for Gramm matrix, one of its application is for creating the matrix of changing basis. (since this matrix is invertible) your result will be immediate. — SaeidAli, May 19 '17 at 12:52
@janmarqz no, it doesn't. it is just the formula from the textbooks. I want to know whether or not it needs to have $n$ rows and why? — PhysicsMathsLove, May 19 '17 at 13:09

PM. · Accepted Answer · 2017-05-19T20:02:35.410

1

Call $P$ the change of basis matrix of dimension $p \times q$, and $\mathbf{v}_E$ be a vector wrt the E basis and $\mathbf{v}_F$ the vector wrt the F basis.

$\mathbf{v}_E$ is $n \times 1$ so for compatibility of multiplication: $P \,\mathbf{v}_E$, we require $q=n$.

Given $\mathbf{v}_F$ is also $n \times 1$ then $p=n$ so $P$ is $n \times n$

Explaining the last line:

$\mathbf{v}_F \in U$ and must have $n$ components.

The $(i^{th}$ component of $\mathbf{v}_F) = (i^{th} \text{row of P}) \times \,\mathbf{v}_E$

Therefore $P$ must have the same number of rows as $\mathbf{v}_F$ has components.

Therefore $P$ has $n$ rows.

edited May 19 '17 at 20:02

answered May 19 '17 at 14:24

PM.

5,409

Can you explain the last line a bit more? – PhysicsMathsLove May 19 '17 at 17:24
@PhysicsMathsLove see the edit – PM. May 19 '17 at 20:02

score 1 · Answer 2 · answered May 22 '17 at 05:02

If $E=\{e_1,...,e_n\}$ is the first basis and $F=\{f_1,...,f_n\}$ is the second then we can write each $$f_1=P_{11}e_1+\cdots+P_{n1}e_n$$ $$f_2=P_{12}e_1+\cdots+P_{n2}e_n$$ $$\cdots$$ $$f_n=P_{1n}e_1+\cdots+P_{nn}e_n$$ where you need a set of $n\times n$ coefficients $P_{ij}$.

And to warranty linear independence of the $f_i$, it is better that the matrix $$P=[P_{ij}],$$ have determinant non-zero.

Change of basis formula - intuition/is this true?

2 Answers2