# Is T invertible? Compute T^(-1) if it exists.

• Dec 11th 2009, 11:59 AM
Undefdisfigure
Is T invertible? Compute T^(-1) if it exists.
Hi, I'm using the book "Linear Algebra" fourth edition by Stephen H. Friedberg, Arnold J. Insel and Lawrence E. Spence. It's a first level university algebra course (after the college level).

The question is Let T:P2(R)-->R^3 be defined by T(f(x))=(f(-1), f(0), f(1)). Is T invertible? Compute T^(-1) if it exists.

Unfortunately the book and the class notes do far from spoon feed you. I have many other questions but if a few of them are answered maybe I'll figure out how to approach these problems.

Thanks.
• Dec 11th 2009, 04:17 PM
Jameson
Quote:

Originally Posted by Undefdisfigure
Hi, I'm using the book "Linear Algebra" fourth edition by Stephen H. Friedberg, Arnold J. Insel and Lawrence E. Spence. It's a first level university algebra course (after the college level).

The question is Let T:P2(R)-->R^3 be defined by T(f(x))=(f(-1), f(0), f(1)). Is T invertible? Compute T^(-1) if it exists.

Unfortunately the book and the class notes do far from spoon feed you. I have many other questions but if a few of them are answered maybe I'll figure out how to approach these problems.

Thanks.

I am just taking a stab at this since no one has answered yet, but I am not great at Linear Algebra so sorry if I'm off base.

If T:V->W then T^(-1):W->V and T is one-to-one and onto, or a bijection. P2 is spanned by {1,x,x^2} and any f(x) in this form where the leading coefficient is non-zero fails to be one-to-one. So I would say this does not have an inverse.

Again, I could be wrong and hopefully someone much smarter than I will chime in soon.
• Dec 11th 2009, 04:47 PM
Jose27
Lagrange's interpolation polynomial gives that \$\displaystyle T\$ is onto, but consider \$\displaystyle f(x)=x(x+1)(x-1)\$ then \$\displaystyle Tf=0\$ and so \$\displaystyle T\$ is not injective.
• Dec 11th 2009, 05:19 PM
Bruno J.
Quote:

Originally Posted by Jose27
Lagrange's interpolation polynomial gives that \$\displaystyle T\$ is onto, but consider \$\displaystyle f(x)=x(x+1)(x-1)\$ then \$\displaystyle Tf=0\$ and so \$\displaystyle T\$ is not injective.

The transformation is from \$\displaystyle P_2(\mathbb{R})\$, not from \$\displaystyle P_3(\mathbb{R})\$.

As you noticed, Lagrange's interpolation formula proves that the map is onto; and since \$\displaystyle \mbox{dim }P_2(\mathbb{R}) = \mbox{dim }\mathbb{R}^3=3\$, the map is also injective and hence invertible.

Another way to prove it would be to calculate the matrix of the transformation with respect to the some basis, and then show that it is invertible. In fact, if you want to calculate the inverse explicitly, you will have to do this.