Prove Property of Projection Matrices

Hey, lets say I have an orthogonal projection matrix $M(n)$ which is n x n so it depends on the number of observations $n$. Lets say I am projecting some vector $z(n)$, a n x 1 vector,  on this space. I want to prove the following inequality holds (or not)? $$||M(n) z(n) || \leq ||M(n+1) z(n+1) ||$$ where $|| \cdot || $is Euclidean Distance. Note here, M(n+1) is M(n) with one new row appended. so its a (n+1) x (n+1) matrix. Similiarly z(n+1) is a (n+1) x 1 vector where we are simply appending a new observation to z(n). So as I add in new observations, changing my orthogonal projection matrix, I am increasing the sum of squared components. Very happy to give $10 for whoever solves this.
Proofs Linear Algebra Matrices
Ben Shapiro Ben Shapiro
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Mathe Mathe
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  • Ben Shapiro Ben Shapiro
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    thanks for the response me be a bit clearer, so $z_{n+1}$ is $z_n$ with just one more element appended onto it, so we are adding new data.

    • Ben Shapiro Ben Shapiro
      0

      so if it was the identity matrix, then we would be adding the square value of the new observation to the sumation increasing it as long its nonzero

    • Mathe Mathe
      0

      A projection matrix has to be squared.

  • Ben Shapiro Ben Shapiro
    0

    maybe I should change my notation, what I am trying to say is z_{n} is a n x 1 vector and z_{n+1} is an (n+1) x 1 vector. let me change that.

    • Mathe Mathe
      0

      A projection matrix has to be squared, because it's a matrix that goes from $\mathbb{R}^m$ to $\mathbb{R}^m $

    • Ben Shapiro Ben Shapiro
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      ok, hopefully that helps, so if M was an identity matrix, we would have ||I(n+1) z(n+1) || = || z(n+1) || = ||z(n) || + z_{n+1}^2

  • Ben Shapiro Ben Shapiro
    0

    am I missing something still?

    • Mathe Mathe
      0

      You are correct in the idea. But the formula is a little bit complicated. I will edit my response.

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