Understanding change of basis and subspaces (problem)

Question

Can you please help me to understand this?

I was given 2 basis (P₂[R] space)

B₁ = {x^2+ x + 1, x, x-1}

B₂ = {x^2 - x + 1, x, 2}

If S ∈ P₂ is a "proper" - "own" subspace (my teacher said that it means the subspace is included but its not the same) and L₁ and L₂ basis of S.

How can i know if the change of matrix from L₁ to B₂ exist? from L₁ to L₂? What would be the size of the matrix?

Can you please help me to understand this? I dont get it :(

Answer 1

If you have one basis $\{b_1, b_2,\ldots,b_n\}$ and you want to convert it into another basis $\{c_1, c_2, c_3,\ldots,c_m\}$, all you have to do is find out how to express each new basis vector in terms of the old basis vectors:

Find the missing coefficients ($\square$). $$c_1 = \square b_1 + \square b_2 + \square b_3 + \ldots + \square b_n$$ $$c_2 = \square b_1 + \square b_2 + \square b_3 + \ldots + \square b_n$$ $$\vdots$$ $$c_m = \square b_1 + \square b_2 + \square b_3 + \ldots + \square b_n$$

Those coefficients will be the entries in a change-of-basis matrix. The matrix will be $m\times n$ if you have $n$ old vectors and $m$ new vectors.

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We can apply this technique to your problem:

$$B_1 = \{x^2 + x +1,\, x,\, x-1\} \qquad B_2 = \{x^2 - x + 1,\, x,\, 2\}$$

We need to express every vector in $B_2$ as a weighted sum of vectors in $B_1$. Notice that:

$$\begin{align*} \color{maroon}{x^2-x+1} &= 1\cdot [\color{teal}{x^2+x+1}]&-2\cdot[\color{teal}{x}] &+0\cdot[\color{teal}{x-1}] \\ \color{maroon}{x} &= 0\cdot [\color{teal}{x^2+x+1}]&+1\cdot[\color{teal}{x}] &+0\cdot[\color{teal}{x-1}] \\ \color{maroon}{2} &= 0\cdot [\color{teal}{x^2+x+1}]&+2\cdot[\color{teal}{x}] &-2\cdot[\color{teal}{x-1}] \\ \end{align*}$$

And so our change of basis matrix is

$$\begin{bmatrix}1 & -2 & 0\\0&1&0\\0&2&-2\end{bmatrix}$$

It transforms any vector written as $a(\color{maroon}{x^2-x+1}) + b(\color{maroon}{x}) + c(\color{maroon}{2})$ into the corresponding vector $a^\prime(\color{teal}{x^2+x+1}) + b^\prime(\color{teal}{x}) + c^\prime(\color{teal}{x-1})$.

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Now the last question. If $S$ is a proper subspace of $P_2$, you will need to know that its basis will have fewer vectors than $P_2$. Because $P_2$ has three basis vectors, a proper subspace $S$ will have one or two basis vectors.

We don't know what those basis vectors in $L$ will be. However, we know that they will be vectors in $P_2$. We have a basis for $P_2$; therefore, we can write the basis vectors in $L$ as a linear combination of basis vectors in $P_2$. When we write the basis vectors in $L$ as a linear combination of the basis vectors for the main space $P_2$, the matrix of coefficients gives the transformation into that subspace.