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Consider a block matrix $$ V=\begin{bmatrix} 0_{\hat{n}\times n} & I_{\hat{n}\times \hat{n}} & 0_{\hat{n}\times l} & 0_{\hat{n}\times q}\\ A_{p\times n} & 0_{p\times \hat{n}} & B_{p\times l} & 0_{p\times q} \end{bmatrix}, $$ where $0$ and $I$ are zero and identity matrices respectively. Then $Ker(V)$ (basis of null space of $V$) is shown as $$ ker(V)=\begin{bmatrix} V_1 & 0\\ 0 & 0\\ V_3 & 0\\ 0 & I_{q\times q} \end{bmatrix}, $$ where $\begin{bmatrix} V_1 \\ V_3 \end{bmatrix} $ is a basis matrix for $Ker([A\ B])$, and not explicitly writing all the matrix dimensions.

I am trying to figure out how $Ker(V)$ is obtained. My attempt is considering a vector $\tilde{V}= \begin{bmatrix} v_n\\ v_{\hat{n}} \\ v_l\\ v_q \end{bmatrix}, $ of dimension $(n+\hat{n}+l+q)\times 1$ and then using the equality $V\tilde{V}=0$, which is not leading me to the given expression of $Ker(V)$. Kindly clarify my query.

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The way you do it is not wrong. Continue the work you have done so far, you can try considering for what $\tilde{V}$ satisfies $V\tilde{V}$, then you should obtain the following equations:

  1. $Av_n+Bv_l=0$
  2. $Iv_{\hat{n}}=0$
  3. $0v_q=0$

then you are not far away from the answer :)

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