Basis of Nullspace: Linear Algebra & Differential Equations

[dschmidt12]

I am in a linear algebra and differential equations course and have recently been learning how to find a basis for a nullspace, row space, or column space. However, I am EXTREMELY confused by a solution to a question in my textbook. The question asks to find the basis for the null space of a particular matrix. For the question at hand, the determinant is zero, which of course means that it is invertible. However, I'm confused as to why this means that there is no basis for this nullspace -- there is no relevant explanation in my text as far as I can tell. Thank you to anyone who can help solve this problem for me!

 

[VeeEight]

If the determinant is zero then the matrix is not invertible

 

[dschmidt12]

I'm sorry, I meant to write not zero...my mistake; that makes a big difference. :)

 

[aPhilosopher]

This all stems from the fact that matrices distribute over vectors and commute with scalar multiplication: i.e, the fact that matrices are linear transformations.

If Av=0 and Ax=r then A(v + x) = Av + Ax = 0 + r = r.

If I have one solution to Ax = r, then I also have a solution for every element of the nullspace. In this situation, how would I write A^-1? Is it A^-1r = x or A^-1r = x + v? They both get sent to r by A.

This means that a matrix can only be invertible if the nullspace is trivial, in other words, if only zero gets sent to zero.

 

[HallsofIvy]

"Nullspace of A" is, by definition, the set of vectors, x, such that Ax= 0. If A has an inverse, take $A^{-1}$ on both sides: $A^{-1}Ax= x= A^{-1}0= 0$.

That is, if A is invertible, Ax= 0 only for x= 0.

It is true, then that if A is invertible, it is one-to-one.