# Matrix decomposition when one is a linear combination of the others

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?

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$$M=\begin{bmatrix}2&1\\1&1\end{bmatrix}=1\times A+0\times B+0\times C+1\times D.$$
• 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 $3 \ne 4$ – user231343 Oct 14 '18 at 16:01
• 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. – José Carlos Santos Oct 14 '18 at 16:11
• 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 $B$. – José Carlos Santos Oct 14 '18 at 16:30
• @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. – peterwhy Oct 14 '18 at 16:41