Linear Mathematics | Part IB, 2003

(a) Let A=(aij)A=\left(a_{i j}\right) be an m×nm \times n matrix and for each knk \leqslant n let AkA_{k} be the m×km \times k matrix formed by the first kk columns of AA. Suppose that n>mn>m. Explain why the nullity of AA is non-zero. Prove that if kk is minimal such that AkA_{k} has non-zero nullity, then the nullity of AkA_{k} is 1 .

(b) Suppose that no column of AA consists entirely of zeros. Deduce from (a) that there exist scalars b1,,bkb_{1}, \ldots, b_{k} (where kk is defined as in (a)) such that j=1kaijbj=0\sum_{j=1}^{k} a_{i j} b_{j}=0 for every imi \leqslant m, but whenever λ1,,λk\lambda_{1}, \ldots, \lambda_{k} are distinct real numbers there is some imi \leqslant m such that j=1kaijλjbj0\sum_{j=1}^{k} a_{i j} \lambda_{j} b_{j} \neq 0.

(c) Now let v1,v2,,vm\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{m} and w1,w2,,wm\mathbf{w}_{1}, \mathbf{w}_{2}, \ldots, \mathbf{w}_{m} be bases for the same real mm dimensional vector space. Let λ1,λ2,,λn\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n} be distinct real numbers such that for every jj the vectors v1+λjw1,,vm+λjwm\mathbf{v}_{1}+\lambda_{j} \mathbf{w}_{1}, \ldots, \mathbf{v}_{m}+\lambda_{j} \mathbf{w}_{m} are linearly dependent. For each jj, let a1j,,amja_{1 j}, \ldots, a_{m j} be scalars, not all zero, such that i=1maij(vi+λjwi)=0\sum_{i=1}^{m} a_{i j}\left(\mathbf{v}_{i}+\lambda_{j} \mathbf{w}_{i}\right)=\mathbf{0}. By applying the result of (b) to the matrix (aij)\left(a_{i j}\right), deduce that nmn \leqslant m.

(d) It follows that the vectors v1+λw1,,vm+λwm\mathbf{v}_{1}+\lambda \mathbf{w}_{1}, \ldots, \mathbf{v}_{m}+\lambda \mathbf{w}_{m} are linearly dependent for at most mm values of λ\lambda. Explain briefly how this result can also be proved using determinants.

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