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I am trying to show that every matrix of the form $$ M = \begin{bmatrix} b+c & a \\ b & c \end{bmatrix}$$ can be written as a linear combination of three of the following four matrices. $$ A = \begin{bmatrix} 1 & 0 \\[0.3em] 0 & 0 \end{bmatrix},\ B = \begin{bmatrix} 1 & 1 \\[0.3em] 0 & 0 \end{bmatrix},\ C = \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 0 \end{bmatrix},\ D = \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 1 \end{bmatrix}.$$ So if $M=\alpha A + \beta B + \gamma C + \delta D $ then non of the $\alpha, \beta, \gamma, \delta$ becomes zero! How that's possible when the $[M]_{11}$ is a combination of $[M]_{21}$ and $[M]_{22}$ and then must be three of the four mentioned matrices involved but all coefficients are nonzero?

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Since the matrices $A$, $B$, $C$, and $D$ are linearly independent and since the space of all $2\times2$ matrics is $4$-dimensional, every $2\times2$ matrix can be expressed as a linear combination of them.

I don't know what makes you think that three of the four coefficients must be $0$. For instance, if $a=b=c=1$, then$$M=\begin{bmatrix}2&1\\1&1\end{bmatrix}=1\times A+0\times B+0\times C+1\times D.$$

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  • $\begingroup$ No one of them must be zero since M has three independent variables so is of rank 3, but A,B,C,D are linearly independent so are of rank 4; and $3 \ne 4$ $\endgroup$ – user231343 Oct 14 '18 at 16:01
  • $\begingroup$ For example if we choose a,b,c any arbitrary numbers and set alpha to be zero; it forces a=b+c contradictory to our arbitrary choice of a,b,c; and likewise we can't choose beta, gamma, delta always and for any arbitrary a,b,c; so possibility of all alpha, beta, gamma, delta to be nonzero makes r.h.s to be '4-dim' but l.h.s (i.e. M) is of 'dim 3'. $\endgroup$ – user231343 Oct 14 '18 at 16:07
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    $\begingroup$ Yes, the space of all those matrics $M$ is $3$-dimensional. But you cannot choose $\alpha$, $\beta$, $\gamma$, and $\delta$ freely. In fact, we must have $\delta=\alpha+\delta$. And the space$$\left\{\alpha A+\beta B+\gamma C+(\alpha+\beta)D\,\middle|\,\alpha,\beta,\gamma\in F\right\}$$is $3$-dimensional too. So, there is no contradiction. $\endgroup$ – José Carlos Santos Oct 14 '18 at 16:11
  • $\begingroup$ Although they all appear, their coefficients are not independent. As I explained, there is the restriction that the coefficient of $D$ must be the sum of the coefficients of $A$ and $B$. $\endgroup$ – José Carlos Santos Oct 14 '18 at 16:30
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    $\begingroup$ @Edi Consider the set of vectors in the form $\begin{bmatrix}a\\a\end{bmatrix}$ and you want to represent them as $\alpha \begin{bmatrix}1\\0\end{bmatrix} + \beta \begin{bmatrix}0\\1\end{bmatrix}$. The right hand side does not have rank 2, because $\alpha$ and $\beta$ are dependent. $\endgroup$ – peterwhy Oct 14 '18 at 16:41

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