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Thread: Linear Algebra, Projection Matrix

  1. #1
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    Linear Algebra, Projection Matrix

    Question: Suppose that 4 vectors, v1, v2, v3, and v4 are given in $R_{6}$ and we don’t know whether or not these vectors are linearly independent. Explain how you would find the (projection) matrix which projects onto the subspace S = span {v1, v2, v3, v4}.


    Is it correct to say: First find the basis of S to obtain a linearly independent columns of a matrix A that will span S. Then use the basis vectors as the columns of A. To find the projection matrix we use $P=A(A^{T}A)^{-1}A^{T}$.


    Thats all I got in explaining the process. Is this correct?
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  2. #2
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    Re: Linear Algebra, Projection Matrix

    Hey slydez.

    Projection operators have the property P^2 = P.

    You will have to use the idea of a projection with least squares [which is what that formula is discussing].

    Do you understand the concept behind least squares?
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  3. #3
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    Re: Linear Algebra, Projection Matrix

    Hello, thanks for the reply.

    Not sure if it changes anything but in the question, in my first post, the $R_{6}$ should actually be $R^{6}$. In which R is all real number symbol.

    From my understanding of least squares, its just a way to minimize the solution when a there is no solution. In which $A^{T}A\hat{x}=A^{T}b$ and $\hat{x}$ is the least squares. So would I have to find $\hat{x}$, the least squares, and plug into the projection, $proj_{C(A)}b=A\hat{x}$, to find the projection matrix? But I how would I be able to find my $b$ to find my $\hat{x}$ and would my A be the 4x6 matrix (since the $R^{6}$)?
    Last edited by slydez; Nov 27th 2016 at 02:28 PM.
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