Given the original data matrix $$ A=\left[\begin{array}{ccc}8 & 10 & 11 \ 25 & 31 & 33\end{array} ight] $$ and the data points in the new variables defined by the principal components $$ A_{ ext {new }}=\left[\begin{array}{ccc}1.08 & 1.26 & 1.01 \ 26.23 & 32.55 & 34.77\end{array} ight] $$, calculate the covariance matrix in the new variable What is the covariance between the new variables? Ex: 1 Ex: 1.23

Question

Given the original data matrix $$ A=\left[\begin{array}{ccc}8 & 10 & 11 \ 25 & 31 & 33\end{array}ight] $$ and the data points in the new variables defined by the principal components $$ A_{	ext {new }}=\left[\begin{array}{ccc}1.08 & 1.26 & 1.01 \ 26.23 & 32.55 & 34.77\end{array}ight] $$, calculate the covariance matrix in the new variable What is the covariance between the new variables? Ex: 1 Ex: 1.23

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We begin by writing the data points of the new variables as given by

$$
A_{\text{new}} = \begin{bmatrix} 1.08 & 1.26 & 1.01 \ 26.23 & 32.55 & 34.77 \end{bmatrix}.
$$

The columns represent the three observations. Denote the new variables (components) as

$$
x = [1.08,\ 1.26,\ 1.01] \quad \text{and} \quad y = [26.23,\ 32.55,\ 34.77].
$$

Since the covariance matrix is computed from the centered data, we first find the sample means for $$x$$ and $$y$$.

---

**Step 1. Compute the sample means**

For $$x$$:

$$
\bar{x} = \frac{1.08 + 1.26 + 1.01}{3} = \frac{3.35}{3} \approx 1.1167.
$$

For $$y$$:

$$
\bar{y} = \frac{26.23 + 32.55 + 34.77}{3} = \frac{93.55}{3} \approx 31.1833.
$$

---

**Step 2. Compute the deviations from the mean**

For $$x$$, the deviations are:

$$
\begin{aligned}
\Delta x_1 &= 1.08 - 1.1167 \approx -0.0367,\[1mm]
\Delta x_2 &= 1.26 - 1.1167 \approx 0.1433,\[1mm]
\Delta x_3 &= 1.01 - 1.1167 \approx -0.1067.
\end{aligned}
$$

For $$y$$, the deviations are:

$$
\begin{aligned}
\Delta y_1 &= 26.23 - 31.1833 \approx -4.9533,\[1mm]
\Delta y_2 &= 32.55 - 31.1833 \approx 1.3667,\[1mm]
\Delta y_3 &= 34.77 - 31.1833 \approx 3.5867.
\end{aligned}
$$

---

**Step 3. Compute the sample covariance**

With $$n = 3$$ observations, the sample covariance between $$x$$ and $$y$$ is given by

$$
\operatorname{cov}(x,y) = \frac{1}{n-1} \sum_{i=1}^3 (\Delta x_i)(\Delta y_i).
$$

Evaluating the sum:

$$
\begin{aligned}
S &= (\Delta x_1)(\Delta y_1) + (\Delta x_2)(\Delta y_2) + (\Delta x_3)(\Delta y_3)\[1mm]
&\approx (-0.0367)(-4.9533) + (0.1433)(1.3667) + (-0.1067)(3.5867)\[1mm]
&\approx 0.1818 + 0.1957 - 0.3827\[1mm]
&\approx -0.0052.
\end{aligned}
$$

Then,

$$
\operatorname{cov}(x,y) \approx \frac{-0.0052}{2} \approx -0.0026.
$$

---

**Step 4. Interpret the Result**

The covariance between the new variables is very close to zero (approximately $$-0.0026$$). In principal component analysis the new variables (principal components) are uncorrelated and, when computed numerically, rounding errors may yield a very small number near zero.

---

**Final Answer**

The covariance between the new variables is approximately $$0$$.
The covariance between the new variables is approximately 0.

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