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Given the eigenvectors: $$\begin{pmatrix} 2 \\ 3 \end{pmatrix}, \begin{pmatrix}4 \\ 5\end{pmatrix}$$

And this linear combination equation $$ \begin{pmatrix} 2 \\ 4 \end{pmatrix} = c_1 \begin{pmatrix} 2 \\ 3 \end{pmatrix}+c_2\begin{pmatrix}4 \\ 5\end{pmatrix}$$

Solving the equation gives: $$ c_1 = 3, c_2 = -1 $$

Which as far as I know could be written as the vector: $$[v]_u = \begin{pmatrix}3 \\ -1 \end{pmatrix}$$

My question is, in this transition from the vector: \begin{pmatrix}2 \\ 4 \end{pmatrix} to this vector \begin{pmatrix}3 \\ -1 \end{pmatrix}

Is the latter vector $(3, -1)$ in a subspace consisting of the mentioned eigenvectors in the linear combination as the basis vectors for this subspace? I can't help but notice how similar it is to this linear combination in an euclidian space with it's basis vectors: $$\begin{pmatrix} 2 \\ 4 \end{pmatrix} = 2 \begin{pmatrix}1 \\ 0 \end{pmatrix}+4 \begin{pmatrix}0 \\ 1 \end{pmatrix}$$

EDIT:

Let me rephrase the question to a more general one. If we have a linear combination $$ c_1 v_1+c_2v_2 $$

Could one say that this linear combination would create a subspace with the two vectors $v_1$ and $v_2$ as it's basis vectors?

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  • $\begingroup$ I don't understand this question after I read it. Could you rephrase your question? Are you asking if the coordinate vector is in the eigenspace span by the two eigenvectors? $\endgroup$
    – user9464
    Jan 3, 2017 at 16:45
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    $\begingroup$ I'm not sure I completely understand your question, but here are some remarks that might help you. The two vectors given at the beginning of your post give a basis for $\mathbb{R}^2$. This means that if you take any vector $v\in\mathbb{R}^2$, you can write $v$ as a linear combination of those two vectors, and call the coefficients of this linear combination the coordinates of $v$ in your new basis. These coordinates will not in general match the coordinates of $v$ in the usual basis. $\endgroup$
    – 211792
    Jan 3, 2017 at 16:46
  • $\begingroup$ Hi, sorry about that! I've rephrased my question to a more general one now. Thanks for all your answers so far :) $\endgroup$ Jan 3, 2017 at 16:56

2 Answers 2

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If we have a linear combination $$ c_1 v_1+c_2v_2 $$

Could one say that this linear combination would create a subspace with the two vectors $v_1$ and $v_2$ as it's basis vectors?

If $v_1$ and $v_2$ are linearly independent they will form a basis with $\dim(V) = \mathbb{R^2}$ so every linear combination of the form $c_1v_1 + c_2v_2$ will generate a subspace, this is called the $span\{v_1,v_2\}$.

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Note that $\frac{1}{2} \left[ 3 \begin{pmatrix} 4 \\ 5 \end{pmatrix} - 5 \begin{pmatrix} 2 \\ 3 \end{pmatrix} \right] = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, so the space with basis vectors $\Bigg \lbrace \begin{pmatrix} 4 \\ 5 \end{pmatrix} , \begin{pmatrix} 2 \\ 3 \end{pmatrix} \Bigg\rbrace$ is in fact $\mathbb{R}^2$. Hence every vector will be in the subspace with basis vectors the eigenvectors.

In general, $n$ linearly independent vectors will form a basis for $\mathbb{R}^n$, and in this case of $n = 2$ you do indeed have linearly independent vectors.

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