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 $A$. Add to solve later

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. Add to solve later

More from my site

You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra Solve Linear Recurrence Relation Using Linear Algebra (Eigenvalues and Eigenvectors)

Let $V$ be a real vector space of all real sequences $(a_i)_{i=1}^{\infty}=(a_1, a_2, \dots).$ Let $U$ be the subspace of...

Close