# If $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ inconsistent, then $(A+B)\vec{x}=\vec{b}$ inconsistent?

The matrices $A$ and $B$ are of $n \times m$. Suppose there is a $\vec{b}$ such that both $A\vec{x}=B\vec{x}=\vec{b}$ have infinitely many solutions. Is it true that $(A+B)\vec{x}=\vec{b}$ also has infinitely many solutions?

I tried to prove this by considering $A\vec{x}=B\vec{x}=\vec{b}$. So $(A+B)\frac{\vec{x}}{2}=\vec{b}$. Let $\vec{y}=\frac{\vec{x}}{2}$, then $(A+B)\vec{y}=\vec{b}$. Since $\vec{y} \in \mathbb{R}$ and $\vec{x}$ has many solutions, $\vec{y}$ also has infinitely many solutions. Hence, it is true that $(A+B)\vec{x}=\vec{b}$ also has infinitely many solutions.

So far, has what I have done proved the claim that $(A+B)\vec{x}=\vec{b}$ also has infinitely many solutions?

Now, if I take it a little further to make the problem as suppose there is a $\vec{b}$ such that both $A\vec{x}=\vec{b}=B\vec{x}=\vec{b}$ are inconsistent. Is it true that $(A+B)\vec{x}=\vec{b}$ is also inconsistent?

I wanted to do the same as the previous one but it doesn't sound very right to still say $A\vec{x}=B\vec{x}=\vec{b}$ since the system isn't consistent and so there isn't an $\vec{x}$ to make the equation well-defined. If this is the case, what else can I do to test if the claim is true or false?

• The proof in the second paragraph is wrong. We know that infinitely many $x$ exist with $Ax = b$, and infinitely many $z$ exist with $Bz = b$. But that doesnt means that the intersection of these $x$s and $z$s (the solutions to $Ax = Bx = b$) is also infinite. (As in Yuval's example below, it could be that $\{x: Ax=b\} \cap \{z: Bz=b\} = \{0\}$ so $|\{x: Ax=b\} \cap \{z: Bz=b\}| < \infty$.) – TMM Sep 24 '11 at 18:21
• Consider $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$, $b = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$. – Yuval Filmus Sep 24 '11 at 18:23
• @YuvalFilmus: Wouldn't it be $\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0\\ 5 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0\\ 5 \end{bmatrix}=\begin{bmatrix} 0\\ 5 \end{bmatrix}$ $$\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0\\ 5 \end{bmatrix} \neq \begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0\\ 5 \end{bmatrix}$$ Wouldn't the $\vec{x}$ need to be a common one that satisfies the 2 equations? – xenon Sep 24 '11 at 18:29
• @ThijsLaarhoven: So when it says $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$, it doesn't mean that when the equations are are evaluated, the $\vec{x}$ must be the same value in the 2 equations? – xenon Sep 24 '11 at 18:40
• @xEnOn: If the statement is "Suppose there is a $b$ such that both $Ax=b$ and $Bx=b$ have infinitely many solutions." as above, then that should be interpreted as that the (seperate) sets of solutions to $Ax = b$ and $Bx = b$ are both infinite. – TMM Sep 24 '11 at 18:44

Suppose $A$ and $B$ are matrices, and $\vec{b}$ is such that the systems of linear equations $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ have infinitely many solutions in common. Is it true that $(A+B)\vec{t}=\vec{b}$ has infinitely many solutions?

If that's the case, then your argument works: for each common solution $\vec{x}$ to the original two systems, $\frac{1}{2}\vec x$ is a solution to $(A+B)\vec{x}=\vec{b}$; since $\frac{1}{2}\vec{x} = \frac{1}{2}\vec{x'}$ if and only if $\vec{x}=\vec{x'}$, it follows that the latter system has infinitely many solutions as well.

But you need the solutions to be common solutions, as Yuval's example shows.

The second question is false in two interpretations:

Suppose $A$ and $B$ are matrices, and $\vec{b}$ is such that the systems of linear equations $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ have no solutions in common. Is it true that $(A+B)\vec{x}=\vec{b}$ has no solutions?

and

Suppose that $A$ and $B$ are matrices, and $\vec{b}$ is such that the systems of linear equations $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ each has no solutions. Is it true that $(A+B)\vec{x}=\vec{b}$ has no solutions?

This is false; note that any pair of matrices $A$ and $B$ and vector $\vec{b}$ that satisfy the second statement will also satisfy the first, so it suffices to find a counterexample to the second statement. The following works: $$A = \left(\begin{array}{cc}1&0\\0&0\end{array}\right),\quad B=\left(\begin{array}{cc}0&0\\0&1\end{array}\right),\quad \vec{b}=\left(\begin{array}{c}1\\1\end{array}\right).$$

Suppose now we tweak it a bit, perhaps; how about the following?

Suppose $A$ and $B$ are matrices, and $\vec{b}$ is such that $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ each has solutions, but there are no solutions in common to both systems. Is it true that $(A+B)\vec{x}=\vec{b}$ has no solutions?

This is still not true. Take $$A=\left(\begin{array}{crc} 1&-1&0\\ 0&0&1 \end{array}\right),\quad B=\left(\begin{array}{ccr} 1&0&0\\ 0&1&-1 \end{array}\right),\quad \vec{b}=\left(\begin{array}{c}1\\1\end{array}\right).$$ Then $A\vec{x}=\vec{b}$ has solutions: $\vec{x}=(1,0,1)^T$ is a solution. $B\vec{x}=\vec{b}$ also has solutions: $\vec{x}=(1,1,0)^T$ is a solution.

But $A\vec{x}=\vec{b}$ and $B\vec{x}=\vec{b}$ have no solutions in common: if $\vec{x}=(x,y,z)^T$ were a solution, then you would need $x-y=1$, $z=1$, $x=1$, and $y-z=1$. But from $x=z=1$ and $x-y=0$, we get $y=0$; and from $z=1$ and $y-z=1$ we get $y=2$.

However, $$(A+B)\vec{x} = \left(\begin{array}{rrr}2 & -1 & 0\\ 0 & 1 & 0 \end{array}\right)\vec{x} = \left(\begin{array}{c}1\\1\end{array}\right)$$ does have solutions: $(1,1,z)^T$ is a solution for all $z$.