Integrate
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Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix. Which of the following must be true?
Some sort of error occurred. Trying to fix in edit. Character limit?
Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix. Which of the following must be true?
i.) If [imath]A\hat{x}=\hat{b}[/imath] is consistent for all [imath] \hat{b} \in \mathbb{R}^m[/imath], then rank(A) = n.
ii.) If [imath]A\hat{x}=0[/imath] has only trivial solution, then rank(A) = n
iii.) If rank(A)=n, then the rows of A form a linearly dependent set
iv.) If rank(A)=n, and A is square matrix, then A invertible
i.) So we have a non-square matrix, and we let m<n we will be guaranteed that we will have one column that spawns a free variable. Guaranteeing that we will be given consistency, as we can make that variable be whatever we need it to for consistency. But with a matrix m>n our matrix is tall, meaning we are not guaranteed a free variable. It is possible we get 0=1 in one of the rows.
ii.) If we have m<n then once again we are guaranteed a spawned free variable. Then we can't have only the trivial solution, as we are guaranteed an arbitrary value which we could morph into other solutions. If m>n then we are not guaranteed a free variable and will not see a free variable at full rank. At full rank for m>n our last leading 1 will need to equal 0 in order to stay consistent. So this one checks off.
iii.) To be linearly dependent, we need to have no free variables. In such case, we could create vectors from
Some sort of error occurred. Trying to fix in edit. Character limit?
Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix. Which of the following must be true?
i.) If [imath]A\hat{x}=\hat{b}[/imath] is consistent for all [imath] \hat{b} \in \mathbb{R}^m[/imath], then rank(A) = n.
ii.) If [imath]A\hat{x}=0[/imath] has only trivial solution, then rank(A) = n
iii.) If rank(A)=n, then the rows of A form a linearly dependent set
iv.) If rank(A)=n, and A is square matrix, then A invertible
i.) So we have a non-square matrix, and we let m<n we will be guaranteed that we will have one column that spawns a free variable. Guaranteeing that we will be given consistency, as we can make that variable be whatever we need it to for consistency. But with a matrix m>n our matrix is tall, meaning we are not guaranteed a free variable. It is possible we get 0=1 in one of the rows.
ii.) If we have m<n then once again we are guaranteed a spawned free variable. Then we can't have only the trivial solution, as we are guaranteed an arbitrary value which we could morph into other solutions. If m>n then we are not guaranteed a free variable and will not see a free variable at full rank. At full rank for m>n our last leading 1 will need to equal 0 in order to stay consistent. So this one checks off.
iii.) To be linearly dependent, we need to have no free variables. In such case, we could create vectors from
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