Linear Algebra

Intersection of Two Null Spaces is Contained in Null Space of Sum of Two Matrices

Problem 311

Let $A$ and $B$ be $n\times n$ matrices. Then prove that
\[\calN(A)\cap \calN(B) \subset \calN(A+B),\] where $\calN(A)$ is the null space (kernel) of the matrix $A$.

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Definition.

Recall that the null space (or kernel) of an $n \times n$ matrix is
\[\calN(A)=\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}\}.\] The null space $\cal(N)$ is a subspace of the $n$-dimensional vector space $\R^n$.

Proof.

Let $\mathbf{x}$ be an arbitrary vector in the intersection $\calN(A)\cap \calN(B)$.
Then the vector $\mathbf{x}$ belongs to both $\calN(A)$ and $\calN(B)$.
Thus, by definition of the null space, we have
\[A\mathbf{x}=\mathbf{0} \text{ and } B\mathbf{x}=\mathbf{0}.\]

If follows from these equalities that we have
\begin{align*}
(A+B)\mathbf{x}=A\mathbf{x}+B\mathbf{x}=\mathbf{0}+\mathbf{0}=\mathbf{0}.
\end{align*}
Hence $\mathbf{x}$ lies in the null space of the matrix $A+B$, that is, $\mathbf{x}\in \calN(A+B)$.

Since $\mathbf{x}$ is an arbitrary element of $\calN(A)\cap \calN(B)$, we have shown the inclusion
\[\calN(A)\cap \calN(B) \subset \calN(A+B),\] as required.

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