Proof the following statement: If A is invertible, then Ax=0 has only the trivia
ID: 1889917 • Letter: P
Question
Proof the following statement:If A is invertible, then Ax=0 has only the trivial solution.
Example given in class:
Proof: Assume A is infertile then there's a unique B so that AB=BA=I
Let b belong R^n be arbitrary. Then let x=Bb which belong R^n.
calculating Ax=A(Bb)=(AB)b=Ib=b
Now we need to show that x is unique. Suppose not, suppose there is at least two, call them x_1 and x_2 with x_1#x_2.
We know Ax_1=b and Ax_2=b
So Ax_1-Ax_2= vector 0
A(x_1-x_2)=0
Then B(A(x_1-x_2))=B0=0
(BA)(x_1-x_2)
I(x_1-x_2)
(x_1-x_2)=0
X_1=x_2 a contradiction
Explanation / Answer
This is true because a matrix is invertible if and only if its column vectors are all linearly independent (If the columns are linearly dependent on the other hand, the matrix will have determinant zero and thus not be invertible) i.e. if A has column vectors v1,...,vn, then those vectors are linearly independent if and only if the equation a1v1+...+anvn·=0 has only the trivial solution a1=...=an=0 This is equivalent to saying Ax=0 has only the trivial solution. To see this, write the linear independence expression in matrix notatio
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.