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Chapter 3: Problem 3

Prove that for any \(m \times n\) matrix \(A, \operatorname{rank}(A)=0\) if and only if \(A\) is the zero matrix.

Short Answer

Expert verified
In summary, we have proven that for any \(m \times n\) matrix A, rank(A) = 0 if and only if A is the zero matrix. We demonstrated this by showing that if rank(A) = 0, then A must be the zero matrix, and conversely, if A is the zero matrix, then its rank is indeed 0.

Step by step solution

01

Prove that if rank(A) = 0, then A is the zero matrix.

Rank of a matrix is defined as the maximum number of linearly independent rows or columns. When rank(A) = 0, it means that there are no linearly independent rows or columns in A. If there are no linearly independent rows or columns in the matrix A, it means that each row (and column) is a linear combination of other rows (and columns) consisting of only zero scalar multiples. In other words, each row (and column) is a product of 0 by some other row (and column). This implies that every entry of A must be 0. Thus, if rank(A) = 0, then A is the zero matrix.
02

Prove that if A is the zero matrix, then rank(A) = 0.

Now we need to prove the other direction of the statement: if A is the zero matrix, then rank(A) = 0. If A is the zero matrix, all its entries are 0. Then, all rows and columns are scalar multiples of the zero vector (multiplying by 0). Since any non-zero scalar multiple of the zero vector is still the zero vector, we cannot find any linearly independent rows or columns. This means that there are no linearly independent rows or columns in A, and thus its rank is 0. Therefore, if A is the zero matrix, then rank(A) = 0. Given the proofs in steps 1 and 2, we have shown that for any \(m \times n\) matrix A, rank(A) = 0 if and only if A is the zero matrix.

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Most popular questions from this chapter

Find the inverse of the row operation "Replace \(R_{i}\) by \(k R_{j}+k^{\prime} R_{i}\left(k^{\prime} \neq 0\right) . "\)

Let \(V\) denote the set of all solutions to the system of linear equations $$ \begin{array}{r} x_{1}-x_{2}+2 x_{4}-3 x_{5}+x_{6}=0 \\ 2 x_{1}-x_{2}-x_{3}+3 x_{4}-4 x_{5}+4 x_{6}=0 . \end{array} $$ (a) Show that \(S=\\{(0,-1,0,1,1,0),(1,0,1,1,1,0)\\}\) is a linearly independent subset of V. (b) Extend \(S\) to a basis for \(\mathrm{V}\).

Let \(u_{1}=(1,2,4), u_{2}=(2,-3,1), u_{3}=(2,1,-1)\) in \(\mathbf{R}^{3}\). Show that \(u_{1}, u_{2}, u_{3}\) are orthogonal, and write \(v\) as a linear combination of \(u_{1}, u_{2}, u_{3},\) where (a) \(v=(7,16,6),(\mathrm{b}) \quad v=(3,5,2)\)

Reduce each of the following matrices to row canonical form: (a) \(A=\left[\begin{array}{rrrrr}2 & 2 & -1 & 6 & 4 \\ 4 & 4 & 1 & 10 & 13 \\\ 8 & 8 & -1 & 26 & 23\end{array}\right]\) (b) \(B=\left[\begin{array}{rrr}5 & -9 & 6 \\ 0 & 2 & 3 \\ 0 & 0 & 7\end{array}\right]\)

Find the dimension and a basis of the general solution \(W\) of each of the following systems: a. \(x_{1}+3 x_{2}+2 x_{3}-x_{4}-x_{5}=0$$2 x_{1}+6 x_{2}+5 x_{3}+x_{4}-x_{5}=0$$5 x_{1}+15 x_{2}+12 x_{3}+x_{4}-3 x_{5}=0\) b. \(2 x_{1}-4 x_{2}+3 x_{3}-x_{4}+2 x_{5}=0$$3 x_{1}-6 x_{2}+5 x_{3}-2 x_{4}+4 x_{5}=0$$5 x_{1}-10 x_{2}+7 x_{3}-3 x_{4}+18 x_{5}=0\)
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