silentpiano Posted February 2, 2011 Share Posted February 2, 2011 Show that if A ε M_{nxn} is nonsingular and t ≠ 0, then tA is nonsingular and (tA)^{-1} = (1/t)A^{-1}. I need to show an intense proof of this statement. Although I can grasp the concept in my head, I am unsure as to the mathematical reasons and theorems that prove this true. Link to comment Share on other sites More sharing options...

Xerxes Posted February 2, 2011 Share Posted February 2, 2011 Show that if A ε M_{nxn} is nonsingular and t ≠ 0, then tA is nonsingular and (tA)^{-1} = (1/t)A^{-1}. I am not going to do this for you, as it really isn't that hard. But I will give a few pointers. Note that since [math]A[/math] is [math]n \times n[/math] then [math]\det(tA) = t^n \det(A)[/math]. So since [math]\det(A) \ne 0,\,\,\,t\,\, \ne 0[/math] then [math]t^n \ne 0 \Rightarrow \det(tA) \ne 0[/math]. But you must prove the premise [math]\det(tA) = t^n \det(A)[/math]. Can you do that? For the second part, namely [math](tA)^{-1} = \frac{1}{t}A^{-1}[/math] you need only to prove that [math] (AB)^{-1} = B^{-1}A^{-1}[/math], remembering that you can treat [math]t[/math] as a [math]1 \times 1[/math] matrix. Recall that 1. [math]AA^{-1}= A^{-1}A =I[/math] 2. matrix algebra is associative 3. if [math] x[/math] is then treated as an element in a commutative ring, here most likely a field, then [math]xA = Ax[/math]. See how you get on 2 Link to comment Share on other sites More sharing options...

Xerxes Posted February 3, 2011 Share Posted February 3, 2011 How disheartening it is to try and help a poster who then doesn't even acknowledge one's efforts, let alone act on them. Worse, it is downright rude. Ah well. 3 Link to comment Share on other sites More sharing options...

imatfaal Posted February 3, 2011 Share Posted February 3, 2011 Know the feeling. Here's a house point. Link to comment Share on other sites More sharing options...

the tree Posted February 3, 2011 Share Posted February 3, 2011 It happens, there isn't much that can be done about it beyond being satisfied that you made a worthy effort. Link to comment Share on other sites More sharing options...

DrRocket Posted February 9, 2011 Share Posted February 9, 2011 I am not going to do this for you, as it really isn't that hard. But I will give a few pointers. Note that since [math]A[/math] is [math]n \times n[/math] then [math]\det(tA) = t^n \det(A)[/math]. So since [math]\det(A) \ne 0,\,\,\,t\,\, \ne 0[/math] then [math]t^n \ne 0 \Rightarrow \det(tA) \ne 0[/math]. But you must prove the premise [math]\det(tA) = t^n \det(A)[/math]. Can you do that? For the second part, namely [math](tA)^{-1} = \frac{1}{t}A^{-1}[/math] you need only to prove that [math] (AB)^{-1} = B^{-1}A^{-1}[/math], remembering that you can treat [math]t[/math] as a [math]1 \times 1[/math] matrix. Recall that 1. [math]AA^{-1}= A^{-1}A =I[/math] 2. matrix algebra is associative 3. if [math] x[/math] is then treated as an element in a commutative ring, here most likely a field, then [math]xA = Ax[/math]. See how you get on It is somewhat simpler to simply note that [math] tA (\frac{1}{t} A^{-1}) = t \frac {1}{t} AA^{-1} = 1 \times I = I[/math] This works for linear operators in general and not just matrices on a finite-dimensional space. Link to comment Share on other sites More sharing options...

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