Suppose that $X$ and $Y$ are two real valued vectors with equal dimensions. Is it true to say $X^TY\le ||X||\cdot||Y||$ where $X^T$ is the transpose of the vector $X$ and $ ||X||$ denotes the 2-norm of the vector $X$ that is : $ ||X||=\sqrt{X^TX}$
2026-04-15 10:29:14.1776248954
Is this true $X^TY\le ||X||\cdot||Y||$?
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Yes, it is true.
We have $X^TY\le |X^TY| \le ||X||.||Y||$
The first inequality is trivial, the second is Cauchy- Schwarz.